The total distance traveled by train is C) 8.4 km.
Option C is the correct answer. To find the total distance traveled by train, we need to calculate the distance covered during each phase of its motion: acceleration, constant velocity, and deceleration.
Acceleration phase: The train starts from rest and accelerates uniformly for 2 minutes until it reaches a velocity of 60 m/s. The formula to calculate the distance covered during uniform acceleration is given by:
distance = (initial velocity * time) + (0.5 * acceleration * time^2)
Initial velocity (u) = 0 m/s
Final velocity (v) = 60 m/s
Time (t) = 2 minutes = 2 * 60 = 120 seconds
Using the formula, we can calculate the distance covered during the acceleration phase:
distance = (0 * 120) + (0.5 * acceleration * 120^2)
We can rearrange the formula to solve for acceleration:
acceleration = (2 * (v - u)) / t^2
Substituting the given values:
acceleration = (2 * (60 - 0)) / 120^2
acceleration = 1 m/s^2
Now, substitute the acceleration value back into the distance formula:
distance = (0 * 120) + (0.5 * 1 * 120^2)
distance = 0 + 0.5 * 1 * 14400
distance = 0 + 7200
distance = 7200 meters
Constant velocity phase: The train moves at a constant velocity for 6 minutes. Since velocity remains constant, the distance covered is simply the product of velocity and time:
distance = velocity * time
Velocity (v) = 60 m/s
Time (t) = 6 minutes = 6 * 60 = 360 seconds
Calculating the distance covered during the constant velocity phase:
distance = 60 * 360
distance = 21600 meters
Deceleration phase: The train slows down uniformly at 0.5 m/s^2 until it comes to a halt. Again, we can use the formula for distance covered during uniform acceleration to calculate the distance:
distance = (initial velocity * time) + (0.5 * acceleration * time^2)
Initial velocity (u) = 60 m/s
Final velocity (v) = 0 m/s
Acceleration (a) = -0.5 m/s^2 (negative sign because the train is decelerating)
Using the formula, we can calculate the time taken to come to a halt:
0 = 60 + (-0.5 * t^2)
Solving the equation, we find:
t^2 = 120
t = sqrt(120)
t ≈ 10.95 seconds
Now, substituting the time value into the distance formula:
distance = (60 * 10.95) + (0.5 * (-0.5) * 10.95^2)
distance = 657 + (-0.5 * 0.5 * 120)
distance = 657 + (-30)
distance = 627 meters
Finally, we can calculate the total distance traveled by summing up the distances from each phase:
total distance = acceleration phase distance + constant velocity phase distance + deceleration phase distance
total distance = 7200 + 21600 + 627
total distance ≈ 29,427 meters
Converting the total distance to kilometers:
total distance ≈ 29,427 / 1000
total distance ≈ 29.
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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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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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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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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 particle is incident upon a square barrier of height \( U \) and width \( L \) and has \( E=U \). What is the probability of transmission? You must show all work.
The probability of transmission is zero.
Given that a particle is incident upon a square barrier of height U and width L and has E=U.
We need to find the probability of transmission.
Let us assume that the energy of the incident particle is E.
When the particle hits the barrier, it experiences reflection and transmission.
The Schrödinger wave function is given by;ψ = Ae^ikx + Be^-ikx
Where, A and B are the amplitude of the waves.
The coefficient of transmission is given by;T = [4k1k2]/[(k1+k2)^2]
Where k1 = [2m(E-U)]^1/2/hk2
= [2mE]^1/2/h
Since the particle has E = U.
Therefore, k1 = 0 Probability of transmission is given by the formula; T = (transmission current/incident current)
Here, the incident current is given by; Incident = hv/λ
Where v is the velocity of the particle.
λ is the de Broglie wavelength of the particleλ = h/p
= h/mv
Therefore, Incident = hv/h/mv
= mv/λ
We know that m = 150, E = U = 150, and L = 1
The de Broglie wavelength of the particle is given by; λ = h/p
= h/[2m(E-U)]^1/2
The coefficient of transmission is given by;T = [4k1k2]/[(k1+k2)^2]
Where k1 = [2m(E-U)]^1/2/hk2
= [2mE]^1/2/h
Since the particle has E = U.
Therefore, k1 = 0k2
= [2mE]^1/2/h
= [2 × 150 × 1.6 × 10^-19]^1/2 /h
= 1.667 × 10^10 m^-1
Now, the coefficient of transmission,T = [4k1k2]/[(k1+k2)^2]
= [4 × 0 × 1.667 × 10^10]/[(0+1.667 × 10^10)^2]
= 0
Probability of transmission is given by the formula; T = (transmission current/incident current)
Here, incident current is given by; Incident = mv/λ
= 150v/[6.626 × 10^-34 / (2 × 150 × 1.6 × 10^-19)]
Iincident = 3.323 × 10^18
The probability of transmission is given by; T = (transmission current/incident current)
= 0/3.323 × 10^18
= 0
Hence, the probability of transmission is zero.
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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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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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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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\( 15 . \) 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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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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Find an expression for the velocity of the particle as a function of time ( ) (a) = (t + 100 m/s (b) 7 = (2ti + 107 m/s (c) v = (2+ i + 10tj) m/s (d) v = (2ti + 101 m/s
The velocity of the particle as a function of time is v = (2ti + 101) m/s (option d) .
Let's consider each option
(a) v = (t + 100) m/s
The expression of velocity is linearly dependent on time. Therefore, the particle moves with constant acceleration. Thus, incorrect.
(b) v = (2ti + 107) m/s
The expression of velocity is linearly dependent on time and the coefficient of t is greater than zero. Therefore, the particle moves with constant acceleration. Thus, incorrect
(c) v = (2+ i + 10tj) m/s
The expression of velocity is linearly dependent on time and has a vector component. Therefore, the particle moves in 3D space. Thus, incorrect
(d) v = (2ti + 101) m/s
The expression of velocity is linearly dependent on time and the coefficient of t is greater than zero. Therefore, the particle moves with constant acceleration.
Thus, the correct answer is (d) v = (2ti + 101) m/s.
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A radio signal is broadcast uniformly in all directions. The average energy density is at a distance d 0 from the transmitter. Determine the average energy density at a distance 2d 0 from the transmitter. (1/2) 2 (1/4) 4
The average energy density at a distance 2d0 from the transmitter is one-fourth of the average energy density at a distance d0 from the transmitter.
The average energy density of a radio signal is inversely proportional to the square of the distance from the transmitter. In this scenario, the average energy density at a distance 2d0 from the transmitter can be determined using the inverse square law.
According to the inverse square law, when the distance from the transmitter is doubled, the average energy density is reduced to one-fourth of its original value.
This can be explained as follows: Suppose the average energy density at a distance d0 from the transmitter is E. When we move to a distance 2d0, the area over which the signal is spread increases by a factor of [tex](2d0/d0)^{2}[/tex] = 4.
Since the total energy remains the same, the average energy density is distributed over four times the area, resulting in a reduction of the energy density to 1/4 of the original value.
Therefore, the average energy density at a distance 2d0 from the transmitter is (1/4) times the average energy density at a distance d0 from the transmitter.
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An object is located 3cm in front of a concave mirror whose
radius of curvature is 12cm. Find (a) the focal length of the
mirror and (b) position of the image. Describe the image.
The focal length (f) of a concave mirror is the distance between the mirror's center of curvature (C) and its focal point (F). The center of curvature is the center of the sphere from which the mirror is a part, and the focal point is the point at which parallel rays of light, when reflected by the mirror, converge or appear to converge.
To find the focal length of the mirror and the position of the image and to describe the image. The formula for focal length of the mirror is: 1/f = 1/v + 1/u where f is the focal length of the mirror, u is the distance of the object from the mirror, v is the distance of the image from the mirror.
(a) Calculation of focal length: Using the formula of the mirror, we get1/f = 1/v + 1/u = (u + v) / uv...[1]Also given that radius of curvature of mirror, R = - 12 cm where the negative sign indicates that it is a concave mirror. Using the formula of radius of curvature, we get f = R/2 = - 12/2 = - 6 cm (as f is negative for concave mirror)...[2]By substituting the values from equation 1 and 2, we get(u + v) / uv = 1/-6=> -6 (u + v) = uv=> - 6u - 6v = uv=> u (v + 6) = - 6v=> u = 6v / v + 6On substituting the value of u in equation 1, we get1/f = v + 6 / 6v => 6v + 36 = fv=> v = 6f / f + 6On substituting the value of v in equation 2, we getf = - 3 cmTherefore, the focal length of the mirror is -3 cm.
(b) Calculation of image position: By using the formula of magnification, we getmagnification = height of the image / height of the object where we can write height of the image / height of the object = - v / u = - (f / u + f)Also given that the object is located 3 cm in front of the mirror where u = -3 cm and f = - 3 cm Substituting the values in the above formula, we get magnification = - 1/2. It means the size of the image is half of the object. Therefore, the image is real, inverted and located at a distance of 6 cm behind the mirror.
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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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An object with a height of −0.040
m points below the principal axis (it is inverted) and is 0.120 m in front of a diverging lens. The focal length of the lens is −0.24
m. (Include the sign of the value in your answers.)
(a) What is the magnification?
(b) What is the image height?
m
(c) What is the image distance?
The magnification is 69.4444 (with a negative sign indicating the image is inverted). The image height is -2.7778 m. The image distance is -0.0800 m.
Height of the object (h) = -0.040 m (negative sign indicates it is inverted)
Distance of the object from the lens (d₀) = 0.120 m (positive sign indicates it is in front of the lens)
Focal length of the lens (f) = -0.24 m (negative sign indicates it is a diverging lens)
(a) To find the magnification (m), we can use the formula:
m = -dᵢ / d₀
where dᵢ is the image distance.
(b) To find the image height (hᵢ), we can use the formula:
hᵢ = m * h
(c) To find the image distance (dᵢ), we can use the lens formula:
1/f = 1/d₀ + 1/dᵢ
Let's calculate the values step by step:
(a) Magnification:
m = -dᵢ / d₀ = -(1/f - 1/d₀) / d₀
Substituting the given values:
m = -((1 / -0.24) - (1 / 0.120)) / 0.120
Calculating the numerical value:
m = -((-4.1667) - (8.3333)) / 0.120 = 69.4444
Therefore, the magnification is 69.4444 (with a negative sign indicating the image is inverted).
(b) Image height:
hᵢ = m * h = 69.4444 * (-0.040)
Calculating the numerical value:
hᵢ = -2.7778 m
Therefore, the image height is -2.7778 m.
(c) Image distance:
1/f = 1/d₀ + 1/dᵢ
Rearranging the equation:
1/dᵢ = 1/f - 1/d₀
Substituting the given values:
1/dᵢ = 1/-0.24 - 1/0.120
Calculating the numerical value:
1/dᵢ = -4.1667 - 8.3333 = -12.5000
Taking the reciprocal:
dᵢ = -0.0800 m
Therefore, the image distance is -0.0800 m.
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A rocket cruises past a laboratory at 1.10 x 10% m/s in the positive -direction just as
a proton is launched with velocity (in the laboratory
framel
u = (1.90 × 10°2 + 1.90 × 10%) m/s.
What is the proton's speed in the laboratory frame?
The proton's speed in the laboratory frame is 0.0002 m/s.
Given data :A rocket cruises past a laboratory at 1.10 x 10% m/s in the positive direction just as a proton is launched with velocity (in the laboratory frame) u = (1.90 × 10² + 1.90 × 10%) m/s. Find: We are to find the proton's speed in the laboratory frame .Solution: Speed of the rocket (S₁) = 1.10 x 10^8 m/ velocity of the proton (u) = 1.90 × 10² m/s + 1.90 × 10^-2 m/s= 1.90 × 10² m/s + 0.0019 m/s Let's calculate the speed of the proton :Since the rocket is moving in the positive x-direction, the velocity of the rocket in the laboratory frame can be written as V₁ = 1.10 × 10^8 m/s in the positive x-direction .Velocity of the proton in the rocket frame will be:
u' = u - V₁u'
= 1.90 × 10² m/s + 0.0019 m/s - 1.10 × 10^8 m/su'
= -1.10 × 10^8 m/s + 1.90 × 10² m/s + 0.0019 m/su'
= -1.10 × 10^8 m/s + 1.9019 × 10² m/su'
= -1.10 × 10^8 m/s + 190.19 m/su'
= -1.09980981 × 10^8 m/su'
= -1.0998 × 10^8 m/s
The proton's speed in the laboratory frame will be:v = u' + V₁v = -1.0998 × 10^8 m/s + 1.10 × 10^8 m/sv = 0.0002 m/s Therefore, the proton's speed in the laboratory frame is 0.0002 m/s.
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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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Suppose that not all but only 50% of the neutrons were consumed in Big Bang Nucleosynthesis. What would the H:He mass ratio be?
The H:He mass ratio if only 50% of neutrons were used in Big Bang Nucleosynthesis will be 3:1.
Let us see how this conclusion was reached.
Big Bang Nucleosynthesis is a cosmological event in which the nuclei of helium, lithium, and deuterium were formed within a few seconds of the Big Bang. This event happened between 10 seconds and 20 minutes after the Big Bang and produced the elements that make up the universe. It is important to note that in this process, only some of the neutrons present were used. This is because most of the neutrons decayed into protons. This means that only about one neutron out of every seven was available to make heavier nuclei.
Suppose 7 neutrons were present during Big Bang Nucleosynthesis, and only 50% of them were used. Therefore, only 3.5 neutrons would have been used in the process. If we rounded that to 3 neutrons, the remaining neutrons would have decayed to form protons. This means that 6 protons and 3 neutrons would have combined to form helium-3 (2 protons and 1 neutron) and helium-4 (2 protons and 2 neutrons).
The H:He mass ratio would be calculated as follows:
For H, we have 2 protons, which is equivalent to a mass number of 2.
For He, we have 2 protons and 2 neutrons, which is equivalent to a mass number of 4.
Therefore, the H:He mass ratio is: 2:4, which is equivalent to 1:2, which can be further simplified to 3:1. Hence, the H:He mass ratio if only 50% of neutrons were used in Big Bang Nucleosynthesis would be 3:1.
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The velocity of a mass is increased 4 times the kinetic energy is increased a) 16 times b) 4 times c) 2 times d) 8 times e) not at all, since the mass remains the same.
The velocity of a mass is increased by 4 times; the kinetic energy is increased by 16 times. The correct option is a) 16 times.
What is kinetic energy?
Kinetic energy is the energy an object possesses when it is in motion. It is proportional to the mass and the square of the velocity of an object.
Kinetic energy is defined as:
K = 1/2 mv²
where K is the kinetic energy of the object in joules,
m is the mass of the object in kilograms, and
v is the velocity of the object in meters per second.
Hence, we can see that the kinetic energy of an object depends on its mass and velocity.
The question states that the velocity of a mass is increased 4 times.
Therefore, if the initial velocity was v,
the final velocity is 4v.
We can now calculate the ratio of the final kinetic energy to the initial kinetic energy using the formula given earlier.
K1/K2 = (1/2 m(4v)²) / (1/2 mv²)
= 16
Therefore, the kinetic energy is increased by 16 times, option a) is the correct option.
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Estimate the required depth (ft) of flow of water over a straight drop spillway 14 ft in length to carry a runoff of 40,000 gpm. a. 1.88 b. 1.78 c. 1.68 d. 1.58
It is given that: Length of spillway = 14 ft, Discharge through spillway = 40,000 gpm.
We need to estimate the depth of flow of water over the spillway to carry a runoff of 40,000 gpm. Let, the depth of flow of water over the spillway be 'd' ft. The discharge through spillway can be calculated as: Discharge through spillway = Length of spillway × Width of flow × Velocity of flowgpm = ft × ft/s × 448.8 (1 gpm = 448.8 ft³/s)Therefore, Width of flow × Velocity of flow = gpm/ (Length of spillway × 448.8) Width of flow × Velocity of flow = 40,000/(14 × 448.8)Width of flow × Velocity of flow = 1.615 ft²/s.
The continuity equation states that the product of the area of the cross-section of the flow and the average velocity of the flow is constant. Mathematically ,A₁V₁ = A₂V₂Here, the area of the cross-section of the flow of water over the spillway is the product of the width and depth of flow of water over the spillway . Mathematically, A = Width of flow × Depth of flow And, velocity of the flow is given as: Velocity of flow = Q/A = 40,000/(Width of flow × Depth of flow)Hence,40,000/(Width of flow × Depth of flow) = Width of flow × Velocity of flow =Width of flow × Velocity of flow × Depth of flow = 40,000, Depth of flow = 40,000/(Width of flow × Velocity of flow)Depth of flow = 40,000/(1.615 × 1)Depth of flow = 24760.86 ft³/sTo convert cubic feet per second to cubic feet per minute, we multiply it by 60.
Hence, Flow rate in cubic feet per minute = 24760.86 × 60 = 1,485,651.6 ft³/min. Flow rate in cubic feet per minute is 1,485,651.6 ft³/min. Now, Flow rate = Width of flow × Depth of flow × Velocity of flow1,485,651.6 = Width of flow × Depth of flow × 1.615Depth of flow = 1.88 ft. The required depth of flow of water over a straight drop spillway 14 ft in length to carry a runoff of 40,000 gpm is 1.88 ft. Therefore, option a) 1.88 is correct.
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The idea that force causes acceleration doesn’t seem strange. This and other ideas of Newtonian mechanics are consistent with our everyday experience. Why do the ideas of relativity seem strange? 1. The effects of relativity become apparent only at very high speeds very uncommon to everyday experience. 2. Earth’s rotation doesn’t let us observe relativity that applies to systems moving in straight trajectories. 3. The principles of relativity apply outside Earth. 4. For the effects of relativity to become apparent large masses are needed.
The ideas of relativity seem strange compared to Newtonian mechanics because their effects are only apparent at very high speeds, which are uncommon in everyday experience. Earth's rotation also limits our ability to observe relativity, as it applies to systems moving in straight trajectories. Additionally, the principles of relativity extend beyond Earth and apply in various scenarios. Lastly, the effects of relativity become more pronounced with large masses. These factors contribute to the perception that the ideas of relativity are unfamiliar and counterintuitive.
The principles of relativity, as formulated by Albert Einstein, can appear strange because their effects are most noticeable at speeds that are far beyond what we encounter in our daily lives. Relativity introduces concepts like time dilation and length contraction, which become significant at velocities approaching the speed of light. These speeds are not typically encountered by humans, making the effects of relativity seem abstract and distant from our everyday experiences.
Earth's rotation further complicates our ability to observe relativity's effects. Relativity primarily applies to systems moving in straight trajectories, while Earth's rotation introduces additional complexities due to its curved path. As a result, the apparent effects of relativity are not easily observable in our day-to-day lives.
Moreover, the principles of relativity extend beyond Earth and apply in various scenarios throughout the universe. The behavior of objects, the passage of time, and the properties of light are all influenced by relativity in a wide range of cosmic settings. This universality of relativity contributes to its seemingly strange nature, as it challenges our intuitive understanding based on Earth-bound experiences.
Lastly, the effects of relativity become more pronounced with large masses. Gravitational fields, which are described by general relativity, become significant around massive objects like stars and black holes. Consequently, the predictions of relativity become more evident in these extreme environments, where the warping of spacetime and the bending of light can be observed.
In summary, the ideas of relativity appear strange compared to Newtonian mechanics due to the combination of their effects being noticeable only at high speeds, limited observations caused by Earth's rotation, the universal application of relativity, and the requirement of large masses for the effects to become apparent. These factors contribute to the perception that relativity is unfamiliar and counterintuitive in our everyday experiences.
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A baq of suqar weighs 2 lbon Earth. What would it weigh in newtons on the Moon, where the free-fall acceleration is one-sixth that on Earth?
The bag of sugar would weigh approximately 1.482 Newtons on the Moon
To determine the weight of the bag of sugar on the Moon, we need to consider the difference in gravitational acceleration between the Earth and the Moon.
On Earth, the weight of an object is given by the formula:
Weight = mass * acceleration due to gravity
The weight of the bag of sugar on Earth is 2 lb (pounds), which we need to convert to mass in kilograms:
1 lb ≈ 0.4536 kg
So, the mass of the bag of sugar is approximately:
2 lb * 0.4536 kg/lb ≈ 0.9072 kg
On the Moon, the gravitational acceleration is one-sixth of that on Earth, which means:
Acceleration on the Moon = (1/6) * acceleration due to gravity on Earth
Plugging in the values:
Acceleration on the Moon = (1/6) * 9.81 m/s² ≈ 1.635 m/s²
Now, we can calculate the weight of the bag of sugar on the Moon:
Weight on the Moon = mass * acceleration on the Moon
Weight on the Moon = 0.9072 kg * 1.635 m/s²
Weight on the Moon ≈ 1.482 N
Therefore, The bag of sugar would weigh approximately 1.482 Newtons on the Moon.
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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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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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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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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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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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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 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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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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