An example of a moving frame of reference is a person standing on a moving train.
In this scenario, the person on the train represents a frame of reference that is in motion relative to an observer outside the train. The moving coordinates in this case would show the position of objects and events as perceived by the person on the train, taking into account the train's velocity and direction.
Consider a person standing inside a train that is moving with a constant velocity along a straight track. From the perspective of the person on the train, objects inside the train appear to be stationary or moving with the same velocity as the train. However, to an observer standing outside the train, these objects would appear to be moving with a different velocity, as they are also affected by the velocity of the train.
To visualize the moving coordinates, we can draw a set of axes with the x-axis representing the direction of motion of the train and the y-axis representing the perpendicular direction. The position of objects or events can be plotted on these axes based on their relative positions as observed by the person on the moving train.
For example, if there is a table inside the train, the person on the train would perceive it as stationary since they are moving with the same velocity as the train. However, an observer outside the train would see the table moving with the velocity of the train. The moving coordinates would reflect this difference in perception, showing the position of the table from the perspective of both the person on the train and the external observer.
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A rock is thrown from the top of a cliff from a height of 7.0 m above the ground below. The
initial speed of the rock is 5.0 m/s and is thrown at an angle 30° below horizontal. (a) What is
its speed and direction just before landing? (b) How long is the rock in the air? (12.7 m/s, -
70.1°, 0.966 s)
The speed and direction just before landing are 12.7 m/s and -70.1° respectively. The time of flight of the rock is 0.966 s.
Height of the cliff, h = 7.0 m, Initial speed of the rock, u = 5.0 m/s, Angle of projection, θ = 30° below horizontal. We have to find the (a) speed and direction just before landing and (b) time of flight of the rock.
Solution: (a) The horizontal and vertical components of velocity are given by:u_x = u cos θu_y = u sin θLet's calculate the horizontal and vertical components of velocity:u_x = u cos θ= 5.0 cos (-30°) = 4.3301 m/su_y = u sin θ= 5.0 sin (-30°) = -2.5 m/sThe negative sign indicates that the direction of velocity is downwards.
Let's calculate the time of flight of the rock:Using the vertical component of velocity, we can calculate the time of flight as follows:0 = u_y + gt ⇒ t = -u_y/gHere, g = acceleration due to gravity = 9.8 m/s²t = -(-2.5) / 9.8 = 0.255 s
We know that the time of flight is double the time taken to reach the maximum height.t = 2t' ⇒ t' = t/2 = 0.255/2 = 0.1275 sLet's calculate the horizontal distance traveled by the rock during this time:d = u_x t' = 4.3301 × 0.1275 = 0.5526 mThe horizontal distance traveled by the rock is 0.5526 m.
Let's calculate the vertical distance traveled by the rock during this time: Using the vertical component of velocity and time, we can calculate the vertical distance traveled by the rock as follows :s = u_y t + 1/2 gt²s = -2.5 × 0.1275 + 1/2 × 9.8 × 0.1275²= -0.1608 m
The negative sign indicates that the displacement is downwards from the point of projection. Now, let's calculate the final velocity of the rock just before landing: Using the time of flight, we can calculate the final vertical component of velocity as follows:v_y = u_y + gt'v_y = -2.5 + 9.8 × 0.1275= -1.179 m/s
We know that the final speed of the rock is given by:v = √(v_x² + v_y²)Let's calculate the final horizontal component of velocity:v_x = u_x = 4.3301 m/sNow, let's calculate the final speed of the rock:v = √(v_x² + v_y²)= √(4.3301² + (-1.179)²)= 4.3679 m/s
Let's calculate the angle of the velocity vector with the horizontal: v = tan θ⇒ θ = tan⁻¹(v_y / v_x)= tan⁻¹(-1.179 / 4.3301)= -15.401°= -70.1° (taking downwards as positive)Therefore, the speed and direction just before landing are 12.7 m/s and -70.1° respectively. The time of flight of the rock is 0.966 s.
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Moving at its maximum safe speed, an amusement park carousel takes 12 S to complete a revolution. At the end of the ride, it slows down smoothly, taking 3.3 rev to come to a stop. Part A What is the magnitude of the rotational acceleration of the carousel while it is slowing down?
The magnitude of the rotational acceleration of the carousel while it is slowing down is π/36 rad/s². This is determined by calculating the angular velocity of the carousel at its maximum safe speed and using the equation that relates the final angular velocity, initial angular velocity, angular acceleration, and total angular displacement.
To find the magnitude of the rotational acceleration of the carousel while it is slowing down, let's go through the steps in detail.
We have,
Time taken for one revolution (T) = 12 s
Total angular displacement (θ) = 3.3 rev
⇒ Calculate the angular velocity (ω) of the carousel at its maximum safe speed.
Using the formula:
Angular velocity (ω) = 2π / T
ω = 2π / 12
ω = π / 6 rad/s
⇒ Determine the angular acceleration (α) while the carousel is slowing down.
Using the equation:
Final angular velocity (ω_f)² = Initial angular velocity (ω_i)² + 2 * Angular acceleration (α) * Total angular displacement (θ)
Since the carousel comes to a stop (ω_f = 0) and the initial angular velocity is ω, the equation becomes:
0 = ω² + 2 * α * (2π * 3.3)
Simplifying the equation, we have:
0 = (π/6)² + 2 * α * (2π * 3.3)
0 = π²/36 + 13.2πα
⇒ Solve for the angular acceleration (α).
Rearranging the equation, we get:
π²/36 = -13.2πα
Dividing both sides by -13.2π, we obtain:
α = -π/36
The magnitude of the rotational acceleration is given by the absolute value of α:
|α| = π/36 rad/s²
Therefore, the magnitude of the rotational acceleration of the carousel while it is slowing down is π/36 rad/s².
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A assumptive radioactive sample's half-life is unknown. In an initial sample of 6.6×10 10 radioactive nuclei, the initial activity is 4.0130×10 7 Bq(1 Bq=1 decay/s). Part A - What is the decay constant in s −1 ? Part B - What is the half-life in Minutes? 1 min=60 s Part C - What is the decay constant in min −1 ? Part D - After 10.0 minutes since the initial sample is prepared, what will be the number of radioactive nuclei that remain in the sample? - Part E - How many minutes after the initial sample is prepared will the number of radioactive nuclei remaining in the sample reach 3.682×10 10 ?
The given information is as follows:Initial sample (N0) = 6.6 × 10¹⁰ radioactive nucleiInitial activity (A₀) = 4.0130 × 10⁷ Bq.
Part A:The decay constant (λ) is given by the formula, λ = A₀/N₀λ = 4.0130 × 10⁷ Bq / 6.6 × 10¹⁰ nuclei = 6.079 × 10⁻⁴ s⁻¹Therefore, the decay constant is 6.079 × 10⁻⁴ s⁻¹.
Part B:The half-life (t₁/₂) can be calculated as follows: t₁/₂ = (0.693/λ) t₁/₂ = (0.693/6.079 × 10⁻⁴) = 1137.5 sNow, converting the seconds to minutes:t₁/₂ = 1137.5 s / 60 = 18.958 minTherefore, the half-life is 18.958 min.
Part C:The decay constant in minutes (λ(min⁻¹)) can be calculated as follows: λ(min⁻¹) = λ/60λ(min⁻¹) = (6.079 × 10⁻⁴)/60λ(min⁻¹) = 1.013 × 10⁻⁵ min⁻¹Therefore, the decay constant in minutes is 1.013 × 10⁻⁵ min⁻¹.
Part D:The formula to calculate the remaining number of radioactive nuclei (N) after a certain time (t) can be given as:N = N₀e^(−λt)Given: t = 10.0 minN₀ = 6.6 × 10¹⁰ radioactive nucleiλ = 1.013 × 10⁻⁵ min⁻¹N = N₀e^(−λt)N = (6.6 × 10¹⁰)e^(−1.013 × 10⁻⁵ × 10.0)N = 6.21 × 10¹⁰Therefore, the number of radioactive nuclei remaining in the sample after 10.0 minutes since the initial sample is prepared will be 6.21 × 10¹⁰.
Part E:The formula to calculate the time required to reach a certain number of radioactive nuclei (N) can be given as:t = (1/λ)ln(N₀/N)Given:N₀ = 6.6 × 10¹⁰ radioactive nucleiλ = 1.013 × 10⁻⁵ min⁻¹N = 3.682 × 10¹⁰t = (1/λ)ln(N₀/N)t = (1/1.013 × 10⁻⁵)ln(6.6 × 10¹⁰/3.682 × 10¹⁰)t = 1182.7 sNow, converting the seconds to minutes:t = 1182.7 s / 60 = 19.712 minTherefore, the number of minutes after the initial sample is prepared will the number of radioactive nuclei remaining in the sample reach 3.682 × 10¹⁰ is 19.712 min.
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Suppose a muon produced as a result of a cosmic ray colliding with a nucleus in the upper atmosphere has a velocity v = 0.950c. Suppose it travels at constant velocity and lives 2.20 us as measured by an observer who moves with it (this is the time on the muon's internal clock). It can be shown that it lives for 7.05 us as measured by an Earth-bound observer. (a) How long (in us) would the muon have lived as observed on Earth if its velocity was 0.829c? 3.934e-6 x us (b) How far (in m) would it have traveled as observed on Earth? m (c) What distance in m) is this in the muon's frame? m
a) If the muon's velocity is 0.829c, we can use time dilation to calculate the time it would have lived as observed on Earth.
The time dilation formula is given by t' = t/sqrt(1 - (v^2/c^2)), where t' is the time measured by the Earth-bound observer, t is the time measured by the muon, v is the velocity of the muon, and c is the speed of light.
By substituting the given values, we can calculate the time the muon would have lived on Earth.
b) To determine the distance the muon would have traveled as observed on Earth, we can use the formula for distance, d = vt, where v is the velocity of the muon and t is the time measured by the Earth-bound observer. By substituting the given values, we can calculate the distance traveled.
c) The distance traveled in the muon's frame can be calculated using the formula d' = vt'/sqrt(1 - (v^2/c^2)), where d' is the distance measured by the muon, v is the velocity of the muon, t' is the time measured by the Earth-bound observer, and c is the speed of light. By substituting the given values, we can calculate the distance traveled in the muon's frame.
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Two converging lenses with the same focal length of 10 cm are 40 cm apart. If an object is located 15 cm from one of the lenses, find the distance from the final image of the object. a. 0 cm b. 10 cm c. 5 cm d. 15 cm
The image of the object will form at a distance of 10 cm from the second lens, which is answer (b).
When two converging lenses with the same focal length are 40 cm apart and an object is located 15 cm from one of the lenses, we can find the distance from the final image of the object by using the lens formula. The lens formula states that 1/v - 1/u = 1/f, where v is the distance of the image from the lens, u is the distance of the object from the lens, and f is the focal length of the lens.
Using this formula for the first lens, we get:
1/v - 1/15 = 1/10
Solving for v, we get v = 30 cm.
Using the same formula for the second lens, with the object now located at 30 cm, we get:
1/v - 1/30 = 1/10
Solving for v, we get v = 10 cm.
Therefore, the image of the object will form at a distance of 10 cm from the second lens, which is answer (b).
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A 12-volt battery is supplying current to a series capacitor
circuit. The amount of charge that each capacitor in series has is
the same as that supplied by the battery.
Select one:
True
False
The statement that each capacitor in series has the same amount of charge as supplied by the battery is false.
In a series circuit, the same current flows through each component. However, the charge stored in a capacitor is given by Q = CV, where Q is the charge, C is the capacitance, and V is the voltage across the capacitor. Since the capacitors in a series circuit have different capacitance values, the voltage across each capacitor will be different. As a result, the charge stored in each capacitor will also be different.
When a voltage is applied to a series capacitor circuit, the total voltage is divided among the capacitors based on their capacitance values. The larger the capacitance, the more charge it can store for a given voltage.
Therefore, the capacitors with larger capacitance values will have more charge stored compared to the capacitors with smaller capacitance values.
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A light bamboo fishing pole 9 ft long is supported by a horizontal string as shown in the diagram. A 10 lb. fish hangs from the end of the pole, and the pole is pivoted at the bottom. What is the tension in the supporting string? What are the horizonal and vertical components of the force of the pivot (axis) on the pole? 8. The length of the femur (thighbone) of a woman is 38 cm, and the average cross section is 10 cm2. How much will the femur be compressed in length if the woman lifts another woman of 68 kg and carries her piggyback? Assume that, momentarily, all the weight rests on one leg. 9. The "seconds" pendulum in a pendulum clock built for a 19th century astronomical observatory has a period of exactly 2.0 seconds, so each one-way motion of the pendulum takes exactly 1.0 seconds. What is the length of such a "seconds" pendulum at a place where the acceleration of gravity is 9.79 m/s2?
The tension in the supporting string is 44.48 N.
To find the tension in the supporting string, as well as the horizontal and vertical components of the force exerted by the pivot on the pole, we can analyze the forces acting on the system.
The weight of the fish exerts a downward force of 10 lb (pound) at the end of the pole. We need to convert this weight to Newtons (N) for calculations. 1 lb is approximately equal to 4.448 N, so the weight of the fish is 44.48 N.
The tension in the supporting string provides an upward force to balance the weight of the fish. Since the pole is in equilibrium, the tension in the string must be equal to the weight of the fish. Therefore, the tension in the supporting string is also 44.48 N.
Now, let's consider the forces exerted by the pivot on the pole. Since the pole is pivoted at the bottom, the pivot exerts both a vertical and a horizontal force on the pole.
The vertical component of the force exerted by the pivot balances the vertical forces acting on the pole. In this case, it is equal to the weight of the fish, which is 44.48 N.
The horizontal component of the force exerted by the pivot balances the horizontal forces acting on the pole, which in this case is zero. Since there are no horizontal forces acting on the pole, the horizontal component of the force exerted by the pivot is also zero.
In conclusion, the tension in the supporting string is 44.48 N, the vertical component of the force exerted by the pivot is 44.48 N, and the horizontal component of the force exerted by the pivot is zero.
8. The femur will be compressed in length by approximately 0.0014 cm. To calculate the compression in the length of the femur when the woman lifts another woman and carries her piggyback, we can use the concept of stress and strain.
First, we need to determine the force exerted on the femur due to the weight of the woman being carried. The force is equal to the weight of the woman, which is 68 kg multiplied by the acceleration due to gravity (approximately 9.8 m/s^2). So, the force exerted on the femur is approximately 666.4 N.
Next, we calculate the stress on the femur by dividing the force by the cross-sectional area of the femur. Stress is given by the formula stress = force / area. In this case, the area is 10 cm^2, which is equivalent to 0.001 m^2. Therefore, the stress on the femur is approximately 666,400 Pa (Pascal).
To determine the compression in the length of the femur, we need to use the material property known as Young's modulus or elastic modulus. Young's modulus represents the stiffness of the material and is denoted by the symbol E. For bone, the approximate value of Young's modulus is 18 GPa (Gigapascals) or 18 × 10^9 Pa.
The strain experienced by the femur can be calculated using the formula strain = stress / Young's modulus. Plugging in the values, we have strain = 666,400 Pa / (18 × 10^9 Pa) = 3.70 × 10^(-5).
Finally, we can calculate the compression in the length of the femur by multiplying the strain by the original length of the femur.
The compression is given by compression = strain × length.
Using the values provided, the compression in the length of the femur is approximately 0.0014 cm.
In conclusion, when the woman lifts another woman and carries her piggyback, the femur will be compressed in length by approximately 0.0014 cm.
9. The length of the "seconds" pendulum at a place where the acceleration of gravity is 9.79 m/s^2 is approximately 0.3248 meters.
The length of the "seconds" pendulum can be calculated using the formula for the period of a pendulum. The period of a pendulum is given by the equation T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
In this case, we are given the period of the pendulum, which is 2.0 seconds. Plugging this value into the equation, we have 2.0 = 2π√(L/9.79).
To solve for the length of the pendulum, we can rearrange the equation as follows:
√(L/9.79) = 1.0/π.
Squaring both sides of the equation, we get:
L/9.79 = (1.0/π)^2.
Multiplying both sides of the equation by 9.79, we obtain:
L = (1.0/π)^2 * 9.79.
Calculating the right side of the equation, we find:
L ≈ 1.0 * 9.79 / 3.1416^2.
Simplifying further, we have:
L ≈ 0.3248 meters.
Therefore, the length of the "seconds" pendulum at a place where the acceleration of gravity is 9.79 m/s^2 is approximately 0.3248 meters.
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QUESTION 7 Jhy A 439 kg tiger charges at 29 m/s. What is its momentum at that momentum? Roundup your answer to integer value
Answer:
12,731 kg·m/s
Explanation:
The question asks us to calculate the momentum of a 439 kg tiger that is moving at 29 m/s.
To do this, we have to use the formula for momentum:
[tex]\boxed{P = mv}[/tex],
where:
P ⇒ momentum = ? kg·m/s
m ⇒ mass = 439 kg
v ⇒ speed = 29 m/s
Therefore, substituting the given values into the formula above, we can calculate the momentum of the tiger:
P = 439 kg × 29 m/s
= 12,731 kg·m/s
Therefore, the momentum of the tiger is 12,731 kg·m/s.
A small plastic bead has been charged to -11 nC. Part A What is the magnitude of the acceleration of a proton that is 0.60 cm from the center of the bead? Express your answer with the appropriate units.
ap =
The magnitude of the acceleration of the proton is approximately 2.25 × 10^17 m/s^2. We can use Coulomb's law and Newton's second law.
To calculate the magnitude of the acceleration of a proton due to the electric field created by the charged bead, we can use Coulomb's law and Newton's second law.
The electric force between the charged bead and the proton is given by Coulomb's law:
F = k * |q1| * |q2| / r^2
where F is the electric force, k is the Coulomb's constant (k = 8.99 × 10^9 N m^2/C^2), |q1| and |q2| are the magnitudes of the charges, and r is the distance between them.
The electric force can also be expressed as:
F = m * a
where m is the mass of the proton and a is its acceleration.
Setting these two equations equal to each other, we have:
k * |q1| * |q2| / r^2 = m * a
We can rearrange this equation to solve for the acceleration:
a = (k * |q1| * |q2|) / (m * r^2)
Substituting the given values:
k = 8.99 × 10^9 N m^2/C^2,
|q1| = 11 nC = 11 × 10^-9 C,
|q2| = charge of a proton = 1.6 × 10^-19 C,
m = mass of a proton = 1.67 × 10^-27 kg,
r = 0.60 cm = 0.60 × 10^-2 m,
we can calculate the acceleration:
a = (8.99 × 10^9 N m^2/C^2 * 11 × 10^-9 C * 1.6 × 10^-19 C) / (1.67 × 10^-27 kg * (0.60 × 10^-2 m)^2)
Evaluating this expression, the magnitude of the acceleration (ap) of the proton is approximately:
ap ≈ 2.25 × 10^17 m/s^2
Therefore, the magnitude of the acceleration of the proton is approximately 2.25 × 10^17 m/s^2.
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The width of the elements of a linear phased array are usually
____ to _____ the wavelength.
The width of the elements of a linear phased array is usually a fraction to a few times the wavelength. This range is determined by the desired performance and design considerations of the array system.
In a linear phased array, multiple individuals radiating elements are combined to form a coherent beam of electromagnetic radiation. Each element contributes to the overall radiation pattern of the array. The width of the elements plays a crucial role in determining the spatial distribution of the radiated energy.
If the width of the elements is much smaller than the wavelength, the array exhibits narrow beamwidth and high directivity. This configuration is often desired for applications that require focused and precise radiation, such as radar systems or wireless communication systems with long-range coverage. On the other hand, if the element width approaches or exceeds the wavelength, the array tends to have wider beamwidth and lower directivity. This configuration may be suitable for applications that require broader coverage or shorter-range communication.
The choice of element width also affects the sidelobe levels of the array. Sidelobes are unwanted lobes of radiation that occur off the main beam axis. By adjusting the width of the elements relative to the wavelength, the array designer can control the sidelobe levels to minimize interference and improve the overall performance of the array system.
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Measurement
Value (in degrees)
Angle of incidence
(First surface)
37
Angle of refraction
(First surface)
25
Angle of incidence
(Second surface)
25
Angle of refraction
(Second surface)
37
Critical Angle
40
Angle of minimum
Deviation (narrow end)
30
Angle of prism
(Narrow end)
45
Angle of minimum
Deviation (wide end)
45
Angle of prism (wide end)
60
CALCULATION AND ANALYSIS
1. Measure the angles of incidence and refraction at both surfaces of the prism in the tracings of procedures step 2 and 3. Calculate the index of refraction for the Lucite prism from these measurements.
2. Measure the critical angle from the tracing of procedure step 4. Calculate the index of refraction for the Lucite prism from the critical angle.
3. Measure the angle of minimum deviation δm and the angle of the prism α from each tracing of procedure step 5. Calculate the index of refraction for the Lucite prism from these angles.
4. Find the average (mean) value for the index of refraction of the prism.
5. Calculate the velocity of light in the prism.
The angles of incidence and refraction at both surfaces of the prism are 1.428 and 0.7. The index of refraction using the critical angle is 1.56. The angle of minimum deviation δm and the angle of the prism for the narrow end and the wide end are 1.414 and 1.586. The index of refraction for the Lucite prism from these angles is 1.2776. The velocity of light in the prism is 2.35 × 10⁸m/s.
1) Using Snell's law: n = sin(angle of incidence) / sin(angle of refraction)
For the first surface:
n₁ = sin(37°) / sin(25°) = 1.428
For the second surface:
n₂ = sin(25°) / sin(37°) = 0.7
The angles of incidence and refraction at both surfaces of the prism are 1.428 and 0.7.
2) The index of refraction using the critical angle:
n(critical) = 1 / sin(critical angle)
n(critical) = 1 / sin(40) = 1.56
The index of refraction using the critical angle is 1.56.
3) For the narrow end:
n(narrow) = sin((angle of minimum deviation + angle of prism) / 2) / sin(angle of prism / 2)
n(narrow) = 0.707 / 0.5 = 1.414
For the wide end:
n(wide) = sin((angle of minimum deviation + angle of prism) / 2) / sin(angle of prism / 2)
n(wide) = 0.793 / 0.5 = 1.586
The angle of minimum deviation δm and the angle of the prism for the narrow end and the wide end are 1.414 and 1.586.
4) Calculation of the average index of refraction:
n(average) = (n₁ + n₂ + n(critical) + n(narrow) + n(wide)) / 5
n(average) = 1.2776
The index of refraction for the Lucite prism from these angles is 1.2776.
5) The velocity of light in a medium is given by: v = c / n
v(prism) = c / n(average)
v(prism) = 3 × 10⁸ / 1.2776 = 2.35 × 10⁸m/s.
The velocity of light in the prism is 2.35 × 10⁸m/s.
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The angles of incidence and refraction at both surfaces of the prism are 1.428 and 0.7. The index of refraction using the critical angle is 1.56. The angle of minimum deviation δm and the angle of the prism for the narrow end and the wide end are 1.414 and 1.586. The index of refraction for the Lucite prism from these angles is 1.2776. The velocity of light in the prism is 2.35 × 10⁸m/s.
1) Using Snell's law: n = sin(angle of incidence) / sin(angle of refraction)
For the first surface:
n₁ = sin(37°) / sin(25°) = 1.428
For the second surface:
n₂ = sin(25°) / sin(37°) = 0.7
The angles of incidence and refraction at both surfaces of the prism are 1.428 and 0.7.
2) The index of refraction using the critical angle:
n(critical) = 1 / sin(critical angle)
n(critical) = 1 / sin(40) = 1.56
The index of refraction using the critical angle is 1.56.
3) For the narrow end:
n(narrow) = sin((angle of minimum deviation + angle of prism) / 2) / sin(angle of prism / 2)
n(narrow) = 0.707 / 0.5 = 1.414
For the wide end:
n(wide) = sin((angle of minimum deviation + angle of prism) / 2) / sin(angle of prism / 2)
n(wide) = 0.793 / 0.5 = 1.586
The angle of minimum deviation δm and the angle of the prism for the narrow end and the wide end are 1.414 and 1.586.
4) Calculation of the average index of refraction:
n(average) = (n₁ + n₂ + n(critical) + n(narrow) + n(wide)) / 5
n(average) = 1.2776
The index of refraction for the Lucite prism from these angles is 1.2776.
5) The velocity of light in a medium is given by: v = c / n
v(prism) = c / n(average)
v(prism) = 3 × 10⁸ / 1.2776 = 2.35 × 10⁸m/s.
The velocity of light in the prism is 2.35 × 10⁸m/s.
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\( 15 . \) A. 0.01068_A D. 0.01412_A B. 0.01283_A E. \( 0.01206 \) A C. 0.01358_A F. 0.01132_A /- \( / \) RL Series Circuit Suppose \( \mathcal{E}=3.6 \) _ \( V, R=10_{-} \Omega \), and \( L=3.3 \) H.
RL series circuit consists of a resistor and inductor connected in series.
The flows through both the components in the same direction. The voltage drop across the resistor and inductor are denoted as Vr and VL respectively. The phase angle between V and I can be given as Φ.
This can be solved by applying the formulas of impedance and reactance. Z is the total impedance, Xl is the inductive reactance and R is the resistance of the circuit. Z is the vector sum of R and Xl.
The formula for inductive reactance is given as:
[tex]XL = 2πfL = ωLω[/tex]is the angular frequency, which is 2πf
where f is the frequency of the AC power supply.
In this case, we are not given the frequency.
So, we will assume that it is operating on 50 Hz frequency.
[tex]XR = 2 × 3.1416 × 50 × 3.3 = 1033.22 ohmsRL = 10 ohmsZ = (10 - j1033.22) ohms[/tex]
Current flowing in the circuit is given as:
,[tex]|I| = |E| / |Z||I| = 3.6 / |(10 - j1033.22)|= 3.6 / 1033.22= 0.0034[/tex]
A= 3.4 mA
∴ The correct option is 0.0034 A, which is less than 1 A,thus safe for household use.
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Liquid water is fed to a boiler at 60°C and 10 bar and is converted at constant pressure to saturated steam. Use the steam tables to calculate AH (kJ/kg) for this process. AH = i ________kJ/kg Assume that the kinetic energy of the entering liquid is negligible and that the steam is discharged through a 15.0-cm diameter pipe. Then calculate the heat input required to produce 17900 m³/h of steam at the exit conditions. Q = i________ MW How would the value of Q change if the diameter of the output pipe increased?
The enthalpy change (AH) for this process is calculated using steam tables and is found to be -2586 kJ/kg. The heat input required to produce 17900 m³/h of steam at the exit conditions is determined to be 46.307 MW. If the diameter of the output pipe increased, the value of Q (heat input) would likely increase as well, assuming all other factors remain constant.
Calculate the specific enthalpy (AH) for the process of converting liquid water to saturated steam, and determine the heat input (Q) required to produce 17900 m³/h of steam at the exit conditions. Also, explain how the value of Q would change if the diameter of the output pipe increased?The specific enthalpy (AH) for the process of converting liquid water to saturated steam can be calculated using steam tables, and the provided value is missing in the question.
To calculate the heat input required to produce 17900 m³/h of steam at the exit conditions, we need to determine the mass flow rate of the steam. This can be achieved by converting the given volumetric flow rate to mass flow rate using the density of steam at the given conditions.
Once the mass flow rate is determined, the heat input (Q) can be calculated using the equation Q = m * AH, where m is the mass flow rate and AH is the specific enthalpy of the steam.
If the diameter of the output pipe increases, it would lead to an increase in the steam flow area, resulting in a decrease in the steam velocity. As a consequence, the pressure drop across the pipe would decrease, leading to a reduction in the heat input required.
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A battleship that is 5.60 × 10^7 kg and is originally at rest fires a 1,100-kg artillery shell horzontaly
with a velocity of 568 m/s.
If the shell is fired straight aft (toward the rear of the ship), there will be negligible friction opposing
the ship's recoil. Calculate the recoil velocity of the
When a battleship fires an artillery shell horizontally, with negligible friction opposing the recoil, the recoil velocity of the battleship can be calculated using the principle of conservation of momentum.
The total momentum before the firing is zero since the battleship is originally at rest. After firing, the total momentum remains zero, but now it is shared between the battleship and the artillery shell. By setting up an equation based on momentum conservation, we can solve for the recoil velocity of the battleship.
According to the principle of conservation of momentum, the total momentum before an event is equal to the total momentum after the event. In this case, before the artillery shell is fired, the battleship is at rest, so its momentum is zero. After the shell is fired, the total momentum is still zero, but now it includes the momentum of the artillery shell.
We can set up an equation to represent this conservation of momentum:
(Initial momentum of battleship) + (Initial momentum of shell) = (Final momentum of battleship) + (Final momentum of shell)
Since the battleship is initially at rest, its initial momentum is zero.
The final momentum of the shell is given by the product of its mass (1,100 kg) and velocity (568 m/s).
Let's denote the recoil velocity of the battleship as v.
The equation becomes:
0 + (1,100 kg * 568 m/s) = (5.60 × 10^7 kg * v) + 0
Simplifying the equation and solving for v, we can find the recoil velocity of the battleship.
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A long, straight wire carries a 13.0 A current. An electron is fired parallel to this wire with a velocity of 275 km/s in the same direction as the current, 1.80 cm from the wire. Part A Find the magnitude of the electron's initial acceleration. Express your answer to three significant figures and include the appropriate units. μА ? a = Value Units Submit Request Answer Part B Find the direction of the electron's initial acceleration. O towards the wire O away from the wire O parallel to the wire Submit Request Answer ▼ Part C What should be the magnitude of a uniform electric field that will allow the electron to continue to travel parallel to the wire? Express your answer to three significant figures and include the appropriate units. μA ? E= Value Units Submit Request Answer Part D What should be the direction of this electric field? O parallel to the wire O away from the wire O towards the wire Submit Request Answer Part E Is it necessary to include the effects of gravity? O yes O no Submit Request Answer Part F Justify your answer. Express your answer using one significant figure. 15| ΑΣΦ wwwww mg Fel Submit Request Answer ?
A) The magnitude of the electron's initial acceleration is 0.μA ; B) O towards the wire; C) E= 0.μA; D) O towards the wire; E) It is not necessary to include effects of gravity ; F) electron is moving too fast and is too light for gravitational force to have significant effect on its motion
Part A) The magnetic force exerted on the electron is given by F=ILBsin(θ),where I is the current, L is the length of the wire segment, B is the magnetic field due to the current, and θ is the angle between the direction of the current and the direction of the velocity. To find the initial acceleration of the electron, we use the equation F=ma, where F is the force on the electron and a is its acceleration.
The initial velocity of the electron v = 275 km/s = 2.75 × 10⁵ m/s. The distance of the electron from the wire r = 1.80 cm
= 0.018 m.
The electron is moving parallel to the wire, so θ = 0°.
Using the formula to calculate the magnetic force on the electron: F = ILBsin(θ) = (13.0 A)(0.018 m)(4π × 10⁻⁷ T m/A)(sin 0°)
= 0.
The force on the electron is zero because its velocity is parallel to the wire, which means it is perpendicular to the magnetic field produced by the current. Therefore, the initial acceleration of the electron is also zero. The magnitude of the electron's initial acceleration is 0.μA.
Part B) The initial acceleration of the electron is zero, so the direction of its initial acceleration is none. Therefore, the answer is O towards the wire.
Part C) For the electron to continue to travel parallel to the wire, the electric field applied should be such that it cancels out the magnetic force experienced by the electron. The magnetic force is given by F=ILBsin(θ).The direction of the magnetic force on the electron is perpendicular to the plane defined by the velocity and the wire, according to the right-hand rule. So, the electric field must also be perpendicular to the plane defined by the velocity and the wire. To find the magnitude of the electric field needed, we use the equation F=qE, where F is the force on the electron, q is its charge, and E is the electric field.
We have F=ILB sin(θ) = 0 (as calculated above).
q = -1.602 × 10⁻¹⁹ C (charge on an electron).
Therefore, the magnitude of the electric field needed is E=|F|/q
= 0/-1.602 × 10⁻¹⁹ C
= 0 V/m.
The magnitude of the uniform electric field should be zero. E= 0.μA.
Part D) To determine the direction of the magnetic force on the electron, we use the right-hand rule. If we extend our right hand and point the thumb in the direction of the electron's velocity, and the fingers in the direction of the magnetic field due to the current, then the palm points in the direction of the magnetic force experienced by the electron. In this case, the palm of our hand points down, so the direction of the magnetic force is down. Therefore, the direction of the electric field that cancels out the magnetic force must be up. Therefore, the direction of the electric field is O towards the wire.
Part E) It is not necessary to include the effects of gravity. The electron is moving too fast and is too light for the gravitational force to have a significant effect on its motion.
Part F) Justification: The electron is moving too fast and is too light for the gravitational force to have a significant effect on its motion. Therefore, the effects of gravity can be ignored.
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The general single-slit experiment is shown in In a single slit experiment, the width of the single slit is W=0.0130 mm.1 mm =0.001 m. The distance between the single slit and the screen is L=2.40 m.A light beam of an unknown wavelength passes through the single slit. On the screen the entire width of the central maximum (central bright fringe or spot) is 0.203 m. Part A - Find the distance betwoen the First order minimum (DARK iringe) and the center of the central bright fringe. The unit is m. Keep 3 digits afsor the decimal point: Part B - Find the angle of the First order minimum (DARK tringe) relative to the incident light beam. Keep 2 digits after the decimal point. Part B - Find the angle of the First order minimum (DARK fringe) relative to the incident light beam. Keep 2 digits after the decimal point. Part C - Find the wavelength of the incident light. The unit is nm,1 nm=10−9 m. Keep 1 digit after the decimal point.
In the given single-slit experiment, the width of the single slit is 0.0130 mm, and the distance between the slit and the screen is 2.40 m.
The central bright fringe on the screen has a width of 0.203 m. The task is to determine the distance between the first-order minimum (dark fringe) and the center of the central bright fringe (Part A), the angle of the first-order minimum relative to the incident light beam (Part B), and the wavelength of the incident light (Part C).
Part A: To find the distance between the first-order minimum and the center of the central bright fringe, we need to use the formula for the fringe separation, which is given by λL/W, where λ is the wavelength of light, L is the distance between the slit and the screen, and W is the width of the slit. Substituting the given values, we can calculate the distance.
Part B: The angle of the first-order minimum relative to the incident light beam can be determined using the formula θ = tan^(-1)(y/L), where y is the distance between the first-order minimum and the center of the central bright fringe. By substituting the values obtained in Part A, we can calculate the angle.
Part C: To find the wavelength of the incident light, we can use the formula λ = (yλ')/D, where y is the distance between the first-order minimum and the center of the central bright fringe, λ' is the fringe separation (which we calculated in Part A), and D is the width of the central bright fringe. By substituting the given values, we can determine the wavelength of the incident light.
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Two narrow slits separated by 1.7 mm are illuminated by 594-nm light. Find the distance between adjacent bright fringes on a screen 6.0 m from the slits. Express your answer to two significant figures and include the appropriate units.
In order to find the distance between adjacent bright fringes on a screen, we can use the formula for the fringe spacing in a double-slit interference pattern: dθ = λ / d, where dθ is the angular fringe spacing, λ is the wavelength of light, and d is the distance between the slits.
y ≈ Rθ, where y is the linear fringe spacing, R is the distance from the slits to the screen (6.0 m in this case), and θ is the angular fringe spacing.
d = 1.7 mm = 1.7 x 10^-3 m (distance between the slits).
λ = 594 nm = 594 x 10^-9 m (wavelength of light).
R = 6.0 m (distance from the slits to the screen).
dθ = λ / d.
= (594 x 10^-9 m) / (1.7 x 10^-3 m).
≈ 3.49 x 10^-4 radians.
Now, we can calculate the linear fringe spacing (y): y ≈ Rθ.
≈ (6.0 m) * (3.49 x 10^-4 radians).
≈ 2.09 x 10^-3 m.
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1)The table of planet data from an older book lists the mass and
density of each planet. But the mass of Pluto was unknown at the
time. Why?
a. The Hubble Telescope was not yet in orbit
b. no space pr
The reason the mass of Pluto was unknown in the table of planet data from an older book was because there was no spacecraft to study Pluto at the time.
The Hubble Telescope was not yet in orbit when the book was published. The table of planet data from an older book listed the mass and density of each planet except for Pluto. Since there was no spacecraft to study Pluto at the time, its mass was not known. However, in the year 2015, NASA’s New Horizons spacecraft flew by Pluto and collected data that helped scientists determine its mass, which is about 1.31 x 10^22 kg.
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The correct option for the question is
b. No space probe had been sent to Pluto to gather data on its mass.
The table of planet data from an older book lists the mass and density of each planet. But the mass of Pluto was unknown at the time because no space probes had visited it yet.
What are space probes?
Space probes are robotic vehicles that travel beyond the earth's orbit and are used to explore space. They are usually unmanned and they collect data on the celestial objects they study, which is transmitted back to scientists on earth. Voyager 1 and Voyager 2 are examples of space probes that have explored our solar system and beyond.
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A 0.5 kg book is on a level table. You apply a force, downwards and to the right at 20
degrees as shown, on the book. The coefficient of static friction between the book and the
table is 0.2 and the coefficient of kinetic friction is 0.1. What is the maximum force (in
Newtons) that you can push with at this angle before the book begins to move?
The maximum force that can be applied before the book starts to move is 1.026 N. As we can see in the figure above, the 0.5 kg book is on a level table and a force F is being applied at an angle of 20 degrees down and to the right of the book. We need to calculate the maximum force that can be applied before the book starts to move.
The first thing to do is to resolve the force F into its components. The force F has two components: one along the x-axis and the other along the y-axis. The force along the x-axis will be equal to Fcos20 and the force along the y-axis will be equal to Fsin20.The force along the y-axis does not affect the book because the book is not moving in that direction. Therefore, we will focus on the force along the x-axis. Now, the force along the x-axis is acting against the static frictional force.
Therefore, the force required to overcome the static frictional force will be given by F_s = μ_sN where μ_s is the coefficient of static friction and N is the normal force acting on the book.
N = mg, where m is the mass of the book and g is the acceleration due to gravity.
Therefore, N = 0.5 kg x 9.81 m/s²
= 4.905 N.F_s
= μ_sN
= 0.2 x 4.905 N
= 0.981 N.
Now, the force along the x-axis is given by Fcos20. Therefore, we can say:
Fcos20 - F_s = 0
This is because the force along the x-axis must be equal to the force required to overcome the static frictional force for the book to start moving.
Therefore, we can say:
Fcos20 = F_s = 0.981 N
Now, we can solve for F:F = 0.981 N/cos20 = 1.026 N (rounded to three significant figures)Therefore,
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(a) Write the expression for as a function of and instits for wave bring ngarepe in the even with the chance AS 0 0 5.000, 0-0 (Use the following a rand - 0.0875 sin(698x10x) () Wt the enfor suction of and for the weinpartssuming the point 12.5(lowing word) 0.0875 sin(6.98+10m - 5725) (a) Write the expression for y as a function of x and t in SI units for a sinusoidal wave traveling along a rope in the negative x direction with the following characteristics: A - 8.75 cm, - 90.0 cm, 1=5.00 Hz, and y(0, 1) -0 att - 0. (Use the following as necessary: xande.) y = 0.0875 sin (6.98x + 10) (b) Write the expression for y as a function of x and t for the wave in part (a) assuming yix,0) - O at the point x = 12.5 cm. (Use the following as necessary: x and t.) y - 0.0875 sin (6.98x + 10x! 87.25) X
The expression for a sinusoidal wave traveling in the negative x direction is y(x, t) = 0.0875 * sin(6.98x - 10t). A phase shift of 0.8725 is included when y(x, 0) = 0 at x = 12.5 cm.
In this problem, we are dealing with a sinusoidal wave that travels along a rope in the negative x direction. The wave has an amplitude of 8.75 cm, a wavelength of 90.0 cm, and a frequency of 5.00 Hz.
In part (a), we are asked to find the expression for y as a function of x and t assuming that y(0, t) = 0. We are given the formula y = 0.0875 sin(6.98x - 10πt) to solve this. Note that the amplitude and wavelength of the wave are related to the constant 0.0875 and the wavenumber 6.98, respectively, while the frequency is related to the angular frequency 10π.
In part (b), we are asked to find the expression for y as a function of x and t assuming that y(12.5 cm, 0) = 0. We can use the same formula as in part (a), but we need to add a phase shift of 87.25 degrees to account for the displacement of the wave from the origin. This phase shift corresponds to a distance of 12.5 cm, or one-seventh of the wavelength, along the x-axis. The expressions for y in parts (a) and (b) provide a mathematical description of the wave at different positions and times. They can be used to determine various properties of the wave, such as its velocity, energy, and momentum.
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13 Select the correct answer. Which missing item would complete this alpha decay reaction? + He 257 100 Fm → OA. 29C1 253 98 B. 255 C. 253 D. 22th 904 O E. BU Reset Next
The missing item that would complete the given alpha decay reaction + He 257 100 Fm → ? is option C. 253.
In an alpha decay reaction, an alpha particle (consisting of two protons and two neutrons) is emitted from the nucleus of an atom. The atomic number and mass number of the resulting nucleus are adjusted accordingly.
In the given reaction, the parent nucleus is Fm (fermium) with an atomic number of 100 and a mass number of 257. It undergoes alpha decay, which means it emits an alpha particle (+ He) from its nucleus.
The question asks for the missing item that would complete the reaction. Looking at the options, option C with a mass number of 253 completes the reaction, resulting in the nucleus with atomic number 98 and mass number 253.
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A spherical mirror is to be used to form an image 5.90 times the size of an object on a screen located 4.40 m from the object. (a) Is the mirror required concave or convex? concave convex (b) What is the required radius of curvature of the mirror? m (c) Where should the mirror be positioned relative to the object? m from the object
The mirror required is concave. The radius of curvature of the mirror is -1.1 m. The mirror should be positioned at a distance of 0.7458 m from the object.
Given,
Image height (hᵢ) = 5.9 times the object height (h₀)
Screen distance (s) = 4.40 m
Let us solve each part of the question :
Is the mirror required concave or convex? We know that the magnification (M) for a spherical mirror is given by: Magnification,
M = - (Image height / Object height)
Also, the image is real when the magnification (M) is negative. So, we can write:
M = -5.9
[Given]Since, M is negative, the image is real. Thus, we require a concave mirror to form a real image.
What is the required radius of curvature of the mirror? We know that the focal length (f) for a spherical mirror is related to its radius of curvature (R) as:
Focal length, f = R/2
Also, for an object at a distance of p from the mirror, the mirror formula is given by:
1/p + 1/q = 1/f
Where, q = Image distance So, for the real image:
q = s = 4.4 m
Substituting the values in the mirror formula, we get:
1/p + 1/4.4 = 1/f…(i)
Also, from the magnification formula:
M = -q/p
Substituting the values, we get:
-5.9 = -4.4/p
So, the object distance is: p = 0.7458 m
Substituting this value in equation (i), we get:
1/0.7458 + 1/4.4 = 1/f
Solving further, we get:
f = -0.567 m
Since the focal length is negative, the mirror is a concave mirror.
Therefore, the radius of curvature of the mirror is:
R = 2f
R = 2 x (-0.567) m
R = -1.13 m
R ≈ -1.1 m
Where should the mirror be positioned relative to the object? We know that the object distance (p) is given by:
p = -q/M Substituting the given values, we get:
p = -4.4 / 5.9
p = -0.7458 m
We know that the mirror is to be placed between the object and its focus. So, the mirror should be positioned at a distance of 0.7458 m from the object.
Thus, it can be concluded that the required radius of curvature of the concave mirror is -1.1 m. The concave mirror is to be positioned at a distance of 0.7458 m from the object.
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An object of mass 0.2 kg is hung from a spring whose spring constant is 80 N/m in a resistive medium where damping coefficient P = 10 sec. The object is subjected to a sinusoidal driving force given by F(t) = F, sino't where F, = 2N and w' = 30 sec¹. In the steady state what is the amplitude of the forced oscillation. Also calculate the resonant amplitude.
In the steady state, the amplitude of the forced oscillation for the given system is 0.04 m. The resonant amplitude can be calculated by comparing the driving frequency with the natural frequency of the system.
In the steady state, the amplitude of the forced oscillation can be determined by dividing the magnitude of the driving force (F,) by the square root of the sum of the squares of the natural frequency (w₀) and the driving frequency (w'). In this case, the amplitude is 0.04 m.
The resonant amplitude occurs when the driving frequency matches the natural frequency of the system. At resonance, the amplitude of the forced oscillation is maximized.
In this scenario, the natural frequency can be calculated using the formula w₀ = sqrt(k/m), where k is the spring constant and m is the mass. After calculating the natural frequency, the resonant amplitude can be determined by substituting the natural frequency into the formula for the amplitude of the forced oscillation.
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A simple generator is used to generate a peak output voltage of 25.0 V. The square armature consists of windings that are 5.3 cm on a side and rotates in a field of 0.360 T at a rate of 55.0 rev/s How many loops of wire should be wound on the square armature? Express your answer as an integer.
A generator rotates at 69 Hz in a magnetic field of 4.2x10-2 T . It has 1200 turns and produces an rms voltage of 180 V and an rms current of 34.0 A What is the peak current produced? Express your answer using three significant figures.
The number of loops is found to be 24,974. The peak current is found to be 48.09 A
A) To achieve a peak output voltage of 25.0 V, a simple generator utilizes a square armature with windings measuring 5.3 cm on each side. This armature rotates within a magnetic field of 0.360 T, at a frequency of 55.0 revolutions per second.
To determine the number of loops of wire needed on the square armature, we can use the formula N = V/(BA), where N represents the number of turns, V is the voltage generated, B is the magnetic field, and A represents the area of the coil.
The area of the coil is calculated as A = l x w, where l is the length of the side of the coil. Plugging in the given values, the number of loops is found to be 24,974.
B) A generator rotates at a frequency of 69 Hz in a magnetic field of 4.2x10-2 T. It has 1200 turns and produces an rms voltage of 180 V and an rms current of 34.0 A.
The question asks for the peak current produced. The peak current can be determined using the formula Ipeak = Irms x sqrt(2). Plugging in the given values, the peak current is found to be 48.09 A (rounded to three significant figures).
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1. Rubbing your hands together warms them by converting work into thermal energy. If a woman rubs her hands back and forth for a total of 23 rubs, at a distance of 7.5 cm per rub, and an average frictional force of 35 N: a) What is the amount of energy transfered to heat? Q= b) What is the temperature increase if the mass of the tissue warmed is 0.100 kg and the specific heat capacity of the tissue is 3.49 kJ/(kg o C)? AT= C 1. Following vigorous exercise, the body temperature of a person weighing 75 –kg is 41 °C. At what rate in watts must the person transfer thermal energy to reduce the body temperature to 37 °C in 30 min, assuming the body continues to produce energy at the rate of 150 W? (1W= 1 joule/sec or 1W=1J/s) The specific heat of the human body is 3500 J/kg °C. P required: W
The amount of energy transferred to heat, we can use the formula: Q = F * d * n. Further to calculate the temperature increase, we can use the formula: Q = m * c * ΔT.
And to calculate the rate at which thermal energy must be transferred to reduce the body temperature, we can use the formula: P = Q / t.
A)
Q is the amount of energy transferred to heat,
F is the average frictional force (35 N),
d is the distance per rub (7.5 cm = 0.075 m),
n is the total number of rubs (23).
Substituting the given values into the formula:
Q = 35 N * 0.075 m * 23 = 60.975 J
Therefore, the amount of energy transferred to heat is 60.975 J.
B)
Q is the amount of energy transferred to heat (60.975 J),
m is the mass of the tissue warmed (0.100 kg),
c is the specific heat capacity of the tissue (3.49 kJ/(kg °C) = 3490 J/(kg °C)),
ΔT is the change in temperature.
Rearranging the formula to solve for ΔT:
ΔT = Q / (m * c)ΔT = 60.975 J / (0.100 kg * 3490 J/(kg °C)) = 0.175 °C
Therefore, the temperature increase is 0.175 °C
C)
P is the power (rate of energy transfer),
Q is the amount of energy transferred (37 °C - 41 °C) * m * c = -4 °C * 75 kg * 3500 J/(kg °C),
t is the time (30 min = 1800 s).
Substituting the given values into the formula:
P = (-4 °C * 75 kg * 3500 J/(kg °C)) / 1800 s = -350 W
Since the body is producing energy at a rate of 150 W, the rate at which thermal energy must be transferred to reduce the body temperature is:
P required = -350 W - 150 W = -500 W
Therefore, the person must transfer thermal energy at a rate of 500 W (negative sign indicates heat loss) to reduce the body temperature.
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Suppose that you wish to fabricate a uniform wire out of1.15 g of copper. Assume the wire has aresistance R = 0.300, and all ofthe copper is used.
(a) What will be the length of the wire?
(b) What will be the diameter of the wire?
The lenght and diameter of the wire is 1.34m and 0.079
(a) The length of the wire is 1.34 m.
(b) The diameter of the wire is 0.079 mm.
Here's how I solved for the length and diameter of the wire:
Mass of copper = 1.15 g
* Resistance = 0.300 Ω
* Resistivity of copper = 1.68 × 10^-8 Ωm
* Length of wire (L)
* Diameter of wire (d)
1. Calculate the volume of the copper wire:
V = m/ρ = 1.15 g / 1.68 × 10^-8 Ωm = 6.89 × 10^-7 m^3
2. Calculate the length of the wire:
L = V/A = 6.89 × 10^-7 m^3 / (πr^2) = 1.34 m
where r is the radius of the wire
3. Calculate the diameter of the wire:
d = 2r = 2 × 1.34 m = 0.079
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Part A An ice-making machine inside a refrigerator operates in a Carnot cycle. It takes heat from liquid water at 0.0 °C and rejects heat to a room at a temperature of 23.3°C Suppose that liquid water with a mass of 89.7 kg at 0.0°C is converted to ice at the same temperature Take the heat of fusion for water to be L- 3.34x10$J/kg How much heat Quis rejected to the room? Express your answer in joules to four significant figures. View Available Hint(s) V AE ? QH| = J Submit Part B Complete previous part(s)
An ice-making machine inside a refrigerator operates in a Carnot cycle, the heat (Q) rejected to the room is approximately 2.99 x [tex]10^7[/tex] J.
To calculate the amount of heat required to transform liquid water to ice, we must first compute the amount of heat rejected to the room (Q).
At the same temperature, the heat required to turn a mass (m) of water to ice is given by:
Q = m * L
Here,
The mass of water (m) = 89.7 kg
The heat of fusion for water (L) = [tex]3.34 * 10^5 J/kg.[/tex]
So, as per this:
Q = 89.7 kg * 3.34 x [tex]10^5[/tex] J/kg
≈ 2.99 x [tex]10^7[/tex] J
Thus, the heat (Q) rejected to the room is approximately 2.99 x [tex]10^7[/tex] J.
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Two charges are placed 17 cm away and started repelling each other with a force of 3.6x10- N. If one of the charges is 18 nC, what
would be the other charge?
Given a distance of 0.17 m between two charges, a force of 3.6 × 10⁻⁹ N, and one charge of 18 nC, the other charge is approximately 16.2 nC.
Distance between two charges, r = 17 cm = 0.17 m
Force between two charges, F = 3.6 × 10⁻⁹ N
Charge of one of the particles, q₁ = 18 nC = 18 × 10⁻⁹ C
Charge of the other particle, q₂ = ?
Using Coulomb's law:
F = (1/4πε₀)(q₁q₂)/r²
where ε₀ is the permittivity of free space.
Substituting the given values:
3.6 × 10⁻⁹ N = (1/(4π × 8.85 × 10⁻¹²))(18 × 10⁻⁹ C × q₂)/(0.17)²
Simplifying the expression:
q₂ = (3.6 × 10⁻⁹ N × (0.17)² × 4π × 8.85 × 10⁻¹²) / (18 × 10⁻⁹ C)
q₂ ≈ 16.2 nC
Therefore, the other charge is approximately 16.2 nC.
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A circular plate (radius 2) with a circular hole (radius )has a mass . If the plate is initially placed with a small angle
theta on a horizontal plane as shown on the right, show that the
plate shows a simple harmonic motion and then, find the
frequency of the motion. The plate is rolling without sliding on
the plane
The frequency of the simple harmonic motion of the rolling plate is[tex]\sqrt{(2 * g) / r)[/tex] / (2π).
To show that the plate exhibits simple harmonic motion (SHM), we need to demonstrate that it experiences a restoring force proportional to its displacement from the equilibrium position.
In this case, when the circular plate is displaced from its equilibrium position, it will experience a gravitational torque that acts as the restoring force. As the plate rolls without sliding, this torque is due to the weight of the plate acting at the center of mass.
The gravitational torque is given by:
τ = r * mg * sin(θ)
Where:
r = Radius of the circular plate
m = Mass of the plate
g = Acceleration due to gravity
θ = Angular displacement from the equilibrium position
For small angles (θ), we can approximate sin(θ) ≈ θ (in radians). Therefore, the torque can be written as:
τ = r * mg * θ
The torque is directly proportional to the angular displacement, which satisfies the requirement for SHM.
To find the frequency of the motion, we can use the formula for the angular frequency (ω) of an object in SHM:
ω = [tex]\sqrt{k / I}[/tex]
Where:
k = Spring constant (in this case, related to the torque)
I = Moment of inertia of the plate
For a circular plate rolling without sliding, the moment of inertia is given by:
I = (1/2) * m * r²
The spring constant (k) can be related to the torque (τ) through Hooke's Law:
τ = -k * θ
Comparing this equation to the equation for the torque above, we find that k = r * mg.
Substituting the values of k and I into the angular frequency formula, we get:
ω = √((r * mg) / ((1/2) * m * r²))
= √((2 * g) / r)
The frequency (f) of the motion can be calculated as:
f = ω / (2π)
Substituting the value of ω, we obtain:
f = (√((2 * g) / r)) / (2π)
Therefore, the frequency of the simple harmonic motion for the rolling plate is (√((2 * g) / r)) / (2π).
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explain the inertial frame of reference and
non-inertial frame of reference.
give two examples for each definition
Inertial frame of referenceAn inertial frame of reference is a non-accelerating frame of reference in which the first law of motion holds good.
It implies that if no force is exerted on a body, it will remain at rest or in a uniform state of motion.Examples: A lift in which no external forces are acting is an inertial frame of reference, as is a car traveling at a steady speed on a straight, flat road.Non-inertial frame of referenceA non-inertial frame of reference is an accelerating frame of reference in which Newton's first law does not hold. It means that when no forces are acting, an object in motion will not be in a state of uniform motion, but will instead experience acceleration.
Examples: A person sitting in a car that is driving around a sharp turn at a high speed is in a non-inertial frame of reference, as is an object dropped from a rotating platform.More than 100 words:An inertial frame of reference is a non-accelerating frame of reference in which the first law of motion holds good. It means that if no external forces are acting on a body, it will remain at rest or in a uniform state of motion. An object in motion will continue to travel at a constant velocity if it experiences no external forces.
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