Electric field strength is the amount of force per unit charge experienced by a test charge in an electric field. It is a vector quantity that can be found by using the following equation: E = F/Q where E represents the electric field strength, F represents the electric force, and Q represents the test charge.
In this problem, we need to find the electric field strength at a point located 1.5 m to the right of the middle charge. We can do this by using the electric field equation for a point charge: E = k * Q / r²where E is the electric field strength, k is the Coulomb constant (8.98755 × 10⁹ N·m²/C²), Q is the charge of the point charge, and r is the distance between the point charge and the point where we want to find the electric field strength. Since we have three point charges in this problem, we need to find the total electric field strength at the point 1.5 m to the right of the middle charge by adding the electric field strengths due to each individual charge. Let's call the middle charge Q2. Then, the electric field strength due to Q2 is given by:E2 = k * Q2 / r²where r is the distance between Q2 and the point 1.5 m to the right of Q2. Since Q2 is located at the midpoint between Q1 and Q3, we can use the Pythagorean theorem to find r:r² = (0.75 m)² + (1.5 m)²r² = 0.5625 m² + 2.25 m²r² = 2.8125 m²r = sqrt(2.8125 m²) = 1.6771 m.
Now we can calculate E2:E2 = k * Q2 / r²E2 = (8.98755 × 10⁹ N·m²/C²) * (5.00 × 10⁻⁶ C) / (1.6771 m)²E2 = 2.6715 N/C Note that the electric field due to Q2 is directed to the left, since Q2 is a negative charge. Now we need to find the electric field due to Q1 and Q3. Since Q1 and Q3 have the same magnitude of charge and are equidistant from the point where we want to find the electric field strength, their electric fields will have the same magnitude and direction. Let's call this magnitude E1:E1 = E3 = k * Q1 / r²where r is the distance between Q1 (or Q3) and the point 1.5 m to the right of Q2. We can again use the Pythagorean theorem to find r:r² = (2.25 m)² + (1.5 m)²r² = 5.0625 m²r = sqrt(5.0625 m²) = 2.25 m Now we can calculate E1 (and E3):E1 = E3 = k * Q1 / r²E1 = E3 = (8.98755 × 10⁹ N·m²/C²) * (5.00 × 10⁻⁶ C) / (2.25 m)²E1 = E3 = 1.1872 N/C Note that the electric field due to Q1 and Q3 is directed to the right, since they are positive charges. Now we can find the total electric field at the point 1.5 m to the right of Q2 by adding the individual electric fields: E total = E1 + E2 + E3Etotal = 1.1872 N/C - 2.6715 N/C + 1.1872 N/CE total = 0.7029 N/C Therefore, the electric field strength at 1.5 m to the right of the middle charge is 0.7029 N/C.
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C.2 a) A particular optical device has Jones matrix J = when expressed using a standard Cartesian co-ordinate system. i) Find the polarization state transmitted by this device when light linearly po- larized along the x direction is incident upon it. [2] ii) Repeat for y polarized incident light. iii) Find the eigenpolarizations and eigenvalues of J. [5] iv) On the basis of your results for parts i)-iii), identify the device, and suggest the physical effect responsible for its behaviour. [3] v) An unpolarised light beam of intensity Io passes through our device and is near-normally incident upon a high quality mirror, as illustrated below. Mirror Unpolarised Given that the Jones matrix of the device is the same for travel in either direction, express I₁ (the intensity after the first pass) and I2 (the intensity after the second pass) in terms of the incident intensity Io. [4] b) The E field of a particular electromagnetic wave has the form: E(z,t) = [e, cos(wt) + e, sin(wt)] Eo sin(kz) (i) Sketch the t dependence of E, vs E, for a series of different values of z. (ii) Sketch also Ez(z) vs Ey(2) for a series of different values of t. [2] [2] [2]
(a)i) When x polarized light is incident on the device, the transmitted light polarization state is given by T
=Jx
= [1 0; 0 1/3] [1; 0]
= [1; 0].ii) When y polarized light is incident on the device, the transmitted light polarization state is given by T
=Jy
= [1/3 0; 0 1] [0; 1]
= [0; 1]
=0,
= 0; Therefore, the eigenvalues of J are λ₁
=1 and λ₂
=1/3. Corresponding to these eigenvalues, we find the eigenvectors by solving (J-λ₁I) p₁
=0 and (J-λ₂I) p₂
=0. Thus, we get: p₁
= [1; 0] and p₂
=[0; 1]. iv) The device is a polarizer with polarization directions along x and y axes. T
= |T|²Io
= 1/3 Io. The reflected beam is also unpolarized, so its intensity is also 1/3 Io.
= 2/3 Io.
=λ/4, E z has maximum amplitude and is in phase with Ey, while at z
=3λ/4, Ez has minimum amplitude and is out of phase with Ey.
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The inductance of a closely packed coil of 420 turns is 11 mH.
Calculate the magnetic flux through the coil when the current is
4.7 mA.
The magnetic flux through the coil when the current is 4.7 mA is approximately 21.714 milliWebers (mWb).
The magnetic flux through a coil can be calculated using the formula:
Φ = L * I
where Φ is the magnetic flux, L is the inductance of the coil, and I is the current passing through the coil.
Given:
Number of turns in the coil (N) = 420
Inductance of the coil (L) = 11 mH = 11 × 10^(-3) H
Current passing through the coil (I) = 4.7 mA = 4.7 × 10^(-3) A
First, we need to calculate the effective number of turns by multiplying the number of turns with the current:
[tex]N_eff = N * IN_eff = 420 * 4.7 × 10^(-3)N_eff = 1.974\\[/tex]
Now, we can calculate the magnetic flux using the formula:
[tex]Φ = L * IΦ = (11 × 10^(-3) H) * (1.974)Φ = 21.714 × 10^(-3) WbΦ = 21.714 mWb\\[/tex]
Therefore, the magnetic flux through the coil when the current is 4.7 mA is approximately 21.714 milliWebers (mWb).
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3. Suppose you have a 9.2 cm diameter fire hose with a 2.4 cm diameter nozzle. Part (a) Calculate the pressure drop due to the Bernoulli effect as water enters the nozzle from the hose at the rate of 40.0 L/s. Take 1.00×10 3 kg/m3 for the density of the water. Part (b) To what maximum height, in meters, above the nozzle can this water rise? (The actual height will be significantly smaller due to air resistance.)
The velocity of water at the nozzle (v2) can be calculated using the volumetric flow rate (Q) and the cross-sectional area of the nozzle.
Part (a) To calculate the pressure drop due to the Bernoulli effect as water enters the nozzle, we can use the Bernoulli equation, which states that the total mechanical energy per unit volume is conserved along a streamline in an ideal fluid flow.
The Bernoulli equation can be written as:
P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2
where P1 and P2 are the pressures at two points along the streamline, ρ is the density of the fluid (given as 1.00×10^3 kg/m^3), v1 and v2 are the velocities of the fluid at those points, g is the acceleration due to gravity (9.8 m/s^2), h1 and h2 are the heights of the fluid at those points.
In this case, we can consider point 1 to be inside the hose just before the nozzle, and point 2 to be inside the nozzle.
Since the water is entering the nozzle from the hose, the velocity of the water (v1) inside the hose is greater than the velocity of the water (v2) inside the nozzle.
We can assume that the height (h1) at point 1 is the same as the height (h2) at point 2, as the water is horizontal and not changing in height.
The pressure at point 1 (P1) is atmospheric pressure, and we need to calculate the pressure drop (ΔP = P1 - P2).
Now, let's calculate the pressure drop due to the Bernoulli effect:
P1 + (1/2)ρv1^2 = P2 + (1/2)ρv2^2
P1 - P2 = (1/2)ρ(v2^2 - v1^2)
We need to find the difference in velocities (v2^2 - v1^2) to determine the pressure drop.
The diameter of the hose (D1) is 9.2 cm, and the diameter of the nozzle (D2) is 2.4 cm.
The velocity of water at the hose (v1) can be calculated using the volumetric flow rate (Q) and the cross-sectional area of the hose (A1):
v1 = Q / A1
The velocity of water at the nozzle (v2) can be calculated using the volumetric flow rate (Q) and the cross-sectional area of the nozzle (A2):
v2 = Q / A2
The cross-sectional areas (A1 and A2) can be determined using the formula for the area of a circle:
A = πr^2
where r is the radius.
Now, let's substitute the values and calculate the pressure drop:
D1 = 9.2 cm = 0.092 m (diameter of the hose)
D2 = 2.4 cm = 0.024 m (diameter of the nozzle)
Q = 40.0 L/s = 0.040 m^3/s (volumetric flow rate)
ρ = 1.00×10^3 kg/m^3 (density of water)
g = 9.8 m/s^2 (acceleration due to gravity)
r1 = D1 / 2 = 0.092 m / 2 = 0.046 m (radius of the hose)
r2 = D2 / 2 = 0.024 m / 2 = 0.012 m (radius of the nozzle)
A1 = πr1^2 = π(0.046 m)^2
A2 = πr2^2 = π(0.012 m)^2
v1 = Q / A1 = 0.040 m^3/s / [π(0.046 m)^2]
v2 = Q / A2 = 0.040 m^3/s / [π(0.012 m)^2]
Now we can calculate v2^2 - v1^2:
v2^2 - v1^2 = [(Q / A2)^2] - [(Q / A1)^2]
Finally, we can calculate the pressure drop:
ΔP = (1/2)ρ(v2^2 - v1^2)
Substitute the values and calculate ΔP.
Part (b) To determine the maximum height above the nozzle that the water can rise, we can use the conservation of mechanical energy.
The potential energy gained by the water as it rises to a height (h) is equal to the pressure drop (ΔP) multiplied by the change in volume (ΔV) due to the expansion of water.
The potential energy gained is given by:
ΔPE = ρghΔV
Since the volume flow rate (Q) is constant, the change in volume (ΔV) is equal to the cross-sectional area of the nozzle (A2) multiplied by the height (h):
ΔV = A2h
Substituting this into the equation, we have:
ΔPE = ρghA2h
Now we can substitute the known values and calculate the maximum height (h) to which the water can rise.
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Even though your internal body temperature stays around 37 degrees C, the temperature of your skin does change. Is the outside of your skin usually at a higher temperature or lower temperature than 37
degrees C? Is the opposite ever true? Explain why.
Changes in core temperature can affect the functioning of organs and bodily systems, leading to health complications.
Even though your internal body temperature stays around 37 degrees C, the temperature of your skin does change. Is the outside of your skin usually at a higher temperature or lower temperature than 37 degrees C?The outside of the skin is usually lower than 37 degrees C, and varies based on environmental conditions. It can range from a few degrees cooler than core temperature in cool conditions to being much warmer than core temperature in hot environments.Is the opposite ever true?The opposite is never true. The outside of the skin cannot be at a higher temperature than core body temperature. The body maintains a temperature range of around 36.5 to 37.5 degrees Celsius, with core temperature being the most constant and sensitive indicator of our body’s temperature.Explanation:Core body temperature is maintained by a homeostatic mechanism regulated by the hypothalamus. When the temperature outside our body changes, the hypothalamus makes the necessary adjustments to keep our internal organs functioning optimally. This is done through actions like shivering or sweating, which are controlled by the autonomic nervous system.Core temperature, on the other hand, is an important measure of health, and changes in core temperature can be a sign of illness. Changes in core temperature can affect the functioning of organs and bodily systems, leading to health complications. This is why doctors often measure body temperature as an indicator of illness.
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A 2.860 kg, 60.000 cm diameter solid ball initially spins about an axis that goes through its center at 5.100 rev/s. A net torque of 1.070 N.m then makes the ball come to a stop. The net work done by the net torque on the ball to make it come to rest, in Joules and to three decimal places, is
The net work done by the net torque on the ball to make it come to rest is approximately -8.422 Joules.
To find the net work done by the net torque on the ball to make it come to rest, we need to use the rotational kinetic energy equation:
K_rot = (1/2) * I * ω²
Where:
K_rot is the rotational kinetic energy
I is the moment of inertia of the ball
ω is the angular velocity
The moment of inertia of a solid sphere rotating about its axis of symmetry can be calculated using the formula:
I = (2/5) * m * r²
Where:
m is the mass of the ball
r is the radius of the ball
Given:
Mass of the ball (m) = 2.860 kg
Diameter of the ball = 60.000 cm
Angular velocity (ω) = 5.100 rev/s
First, we need to convert the diameter of the ball to its radius:
Radius (r) = Diameter / 2 = 60.000 cm / 2 = 30.000 cm = 0.300 m
Now, we can calculate the moment of inertia (I) using the formula:
I = (2/5) * m * r² = (2/5) * 2.860 kg * (0.300 m)²
I = 0.3432 kg·m²
Next, we can calculate the initial rotational kinetic energy (K_rot_initial) using the given angular velocity:
K_rot_initial = (1/2) * I * ω² = (1/2) * 0.3432 kg·m² * (5.100 rev/s)²
K_rot_initial = 8.422 J
Since the net torque causes the ball to come to rest, the final rotational kinetic energy (K_rot_final) is zero. The net work done by the net torque can be calculated as the change in rotational kinetic energy:
Net Work = K_rot_final - K_rot_initial = 0 - 8.422 J
Net Work = -8.422 J
Therefore, the net work done by the net torque on the ball to make it come to rest is approximately -8.422 Joules (to three decimal places).
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An AC source with the effective (rms) voltage of 90 Volt operating at frequency 500 Hz is connected to a 25- resistor, a 12-F capacitor and 30-mH inductor.
Determine:
a. Impedance of the circuit
b. Effective (rms) voltage at resistor, inductor and capacitor
c. Power factor of the circuit d. Instantaneous current, i(t), of the circuit
Answer: It would be A. Impedance of the circuit
Explanation:
The distance between two positive charges is increased. What happens to the potential energy of the system as the distance is increased? The potential energy remains constant. The potential energy decreases. The potential energy increases.
As the distance between two positive charges is increased, the potential energy of the system decreases.
The potential energy between two charges is given by the equation U = k * (q1 * q2) / r, where U is the potential energy, k is the electrostatic constant, q1 and q2 are the magnitudes of the charges, and r is the distance between the charges.Since the charges are positive, their potential energy is positive as well. As the distance between the charges increases (r increases), the denominator of the equation gets larger, resulting in a smaller potential energy. Therefore, the potential energy decreases as the distance between the charges is increased. In summary, the potential energy decreases as the distance between two positive charges is increased.
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A wire 29.0 cm long lies along the z-axis and carries a current of 7.90 A in the +z-direction. The magnetic field is uniform and has components B, = -0.234 T , By = -0.957 T, and B2 = -0.347 T.
a)
Find the x-component of the magnetic force on the wire.
Express your answer in newtons.
b)
Find the y-component of the magnetic force on the wire.
Express your answer in newtons.
c)
Find the z-component of the magnetic force on the wire.
Express your answer in newtons.
d)
What is the magnitude of the net magnetic force on the wire?
Express your answer in newtons.
a) The x-component of the magnetic force on the wire is -0.884 N.
b) The y-component of the magnetic force on the wire is -0.523 N.
c) The z-component of the magnetic force on the wire is 0 N.
d) The magnitude of the net magnetic force on the wire is approximately 1.027 N.
To find the magnetic force on a current-carrying wire, we can use the formula:
F = I × (L x B)
where F is the magnetic force vector, I is the current, L is the length vector of the wire, and B is the magnetic field vector.
a) Finding the x-component of the magnetic force:
The length vector of the wire is given as L = 29.0 cm along the z-axis, which means L = (0, 0, 0.29 m). The magnetic field vector is given as B = (-0.234 T, -0.957 T, -0.347 T).
Using the formula F = I × (L x B), we can calculate the x-component of the magnetic force:
F_x = I × (L x B)_x
= 7.90 A × (0.29 m × (-0.347 T) - 0)
= -0.884 N
Therefore, the x-component of the magnetic force on the wire is -0.884 N.
b) Finding the y-component of the magnetic force:
Using the same formula, we can calculate the y-component of the magnetic force:
F_y = I × (L x B)_y
= 7.90 A × (0.29 m * (-0.234 T) - 0)
= -0.523 N
Therefore, the y-component of the magnetic force on the wire is -0.523 N.
c) Finding the z-component of the magnetic force:
Using the same formula, we can calculate the z-component of the magnetic force:
F_z = I × (L x B)_z
= 7.90 A × (0 - 0)
= 0 N
Therefore, the z-component of the magnetic force on the wire is 0 N.
d) Finding the magnitude of the net magnetic force:
To find the magnitude of the net magnetic force, we can use the formula:
|F| = sqrt(F_x² + F_y² + F_z²)
Plugging in the values, we get:
|F| = √((-0.884 N)² + (-0.523 N)² + (0 N)²)
= √(0.781456 N² + 0.273529 N²)
= √(1.054985 N²)
= 1.027 N
Therefore, the magnitude of the net magnetic force on the wire is approximately 1.027 N.
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3. Three polarizing plates whose planes are parallel are centered on a common axis. The directions of the transmission axes relative to the common vertical direction, as shown below. A linearly polarized beam of light with plane of polarization parallel to the vertical reference direction is incident from the left onto the first disk with intensity Ii =10.0 units (arbitrary). If when θ1=20.0∘,θ2=40.0∘, and θ3=60.0∘, then show that the transmitted intensity is about 6.89 units.
The transmitted intensity through the three polarizing plates is approximately 1.296 units.
To determine the transmitted intensity through the three polarizing plates, considering Malus's Law,
I = Ii × cos²(θ)
Where:
I: transmitted intensity
Ii: incident intensity
θ: angle between the transmission axis of the polarizer and the plane of polarization of the incident light.
Given,
Ii = 10.0 units
θ1 = 20.0°
θ2 = 40.0°
θ3 = 60.0°
Calculate the transmitted intensity through each plate:
I₁ = 10.0 × cos²(20.0°)
I₁ ≈ 10.0 × (0.9397)²
I₁ ≈ 8.821 units
I₂ = 8.821 ×cos²(40.0°)
I₂ ≈ 8.821 ×(0.7660)²
I₂ ≈ 5.184 units
I₃ = 5.184 × cos²(60.0°)
I₃ ≈ 5.184 × (0.5000)²
I₃ ≈ 1.296 units
Therefore, the transmitted intensity is 1.296 units.
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Questions 1. Considering your value for the % difference in the two values, what can you conclude about the slope of the tangent line drawn at a specific point in time on your Height Versus Time graph
The term "% difference" refers to the difference between two values expressed as a percentage of the average of the two values. It can be calculated using the following formula:
% Difference = [(Value 1 - Value 2) / ((Value 1 + Value 2)/2)] x 100
In order to answer this question, we need more information such as the values, the variables and the context of the problem. However, I can provide a general explanation that may be helpful in understanding the concepts mentioned.
The "tangent line" is a straight line that touches a curve at a specific point, without crossing through it. It represents the instantaneous rate of change (or slope) of the curve at that point.
The "Height versus Time graph" is a graph that shows the relationship between the height of an object and the time it takes for the object to fall or rise. Considering the value of the % difference in the two values, we can conclude that the slope of the tangent line drawn at a specific point in time on the Height Versus Time graph will depend on the values of the height and time at that point. If the % difference is small, then the slope of the tangent line will be relatively constant (or flat) at that point. If the % difference is large, then the slope of the tangent line will be more steep or less steep at that point, depending on the direction of the difference and the values of height and time. I hope this helps! If you have any more specific information or questions, please let me know and I'll do my best to assist you.
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14. If a kid’s pool has a volume of 2 m³, how long should a hose
with a radius of 10 cm take to fill it up if it comes out at 0.15
m/s? What is the volume flow rate from the hose?
The volume flow rate from the hose is approximately 0.00471 cubic meters per second.
To calculate the time, it takes to fill the pool and the volume flow rate from the hose, we can use the formulas related to the volume and flow rate of a cylindrical hose.
First, let's convert the radius of the hose from centimeters to meters:
Radius = 10 cm = 0.1 m
The volume of the pool is given as 2 m³. The volume (V) of a cylinder can be calculated using the formula:
V = πr²h
Where:
V is the volume of the cylinder (pool),
π is a mathematical constant approximately equal to 3.14159,
r is the radius of the hose,
and h is the height of the cylinder (pool).
Since we're solving for time, we can rearrange the formula:
h = V / (πr²)
Now we can substitute the given values:
h = 2 m³ / (π(0.1 m)²)
h ≈ 63.66 m
So, the height of the pool is approximately 63.66 meters.
To calculate the time it takes to fill the pool, we can use the formula:
Time = Distance / Speed
The distance is equal to the height of the pool (h), and the speed is given as 0.15 m/s. Therefore:
Time = 63.66 m / 0.15 m/s
Time ≈ 424.4 seconds
So, it would take approximately 424.4 seconds (or about 7 minutes and 4 seconds) to fill the pool.
Next, let's calculate the volume flow rate from the hose. The volume flow rate (Q) is given by the formula:
Q = A * V
Where:
Q is the volume flow rate,
A is the cross-sectional area of the hose,
and V is the velocity of the water coming out of the hose.
The cross-sectional area (A) of a cylinder is given by:
A = πr²
Substituting the values:
A = π(0.1 m)²
A ≈ 0.03142 m²
Now we can calculate the volume flow rate:
Q = 0.03142 m² * 0.15 m/s
Q ≈ 0.00471 m³/s
Therefore, the volume flow rate from the hose is approximately 0.00471 cubic meters per second.
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An ice dancer with her arms stretched out starts into a spin with an angular velocity of 2.2 rad/s. Her moment of inertia with her arms stretched out is 2.74kg m? What is the difference in her rotational kinetic energy when she pulls in her arms to make her moment of inertia 1.54 kg m2?
The difference in rotational kinetic energy when the ice dancer pulls in her arms from a moment of inertia of 2.74 kg m² to 1.54 kg m² is 0.998 Joules.
When the ice dancer pulls in her arms, her moment of inertia decreases, resulting in a change in rotational kinetic energy. The formula for the difference in rotational kinetic energy (ΔK) is given by ΔK = ½ * (I₂ - I₁) * (ω₂² - ω₁²), where I₁ and I₂ are the initial and final moments of inertia, and ω₁ and ω₂ are the initial and final angular velocities.
Given I₁ = 2.74 kg m², I₂ = 1.54 kg m², and ω₁ = 2.2 rad/s, we can calculate ω₂ using the conservation of angular momentum, I₁ * ω₁ = I₂ * ω₂. Solving for ω₂ gives ω₂ = (I₁ * ω₁) / I₂.
Substituting the values into the formula for ΔK, we have ΔK = ½ * (I₂ - I₁) * [(I₁ * ω₁ / I₂)² - ω₁²].
Performing the calculations, we find ΔK ≈ 0.998 Joules. This means that when the ice dancer pulls in her arms, the rotational kinetic energy decreases by approximately 0.998 Joules.
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8 of 11 Newton's Law of Cooling states that the temperature T of an object at any time t, in minutes, can be described by the equation T = Ts + (To-Ts)e-kt, where Ts is the temperature of the surrounding environment, To is the initial temperature of the object, and k is the cooling rate. What is the cooling rate of an object if the initial temperature was 110° C, the surrounding environment temperature was 10° C, and it took 25 minutes to cool down to 35° C. Round your result to 3 decimal places. k = 0.054 k = 0.055 k = 0.057 k = 0.400
The cooling rate of the object is 0.054.
Let's find the cooling rate (k) of an object using the given information. Ts = 10 °CTo = 110 °CT1 = 35 °Ct2 = 25 minutes. Now, the given formula is T = Ts + (To - Ts) e ^ -kt. Here, we know that the temperature drops from 110°C to 35°C, which is 75°C in 25 minutes. Now, we will substitute the values in the formula as follows:35 = 10 + (110 - 10) e ^ (-k × 25) => (35 - 10) / 100 = e ^ (-k × 25) => 25 / 100 = k × 25 => k = 0.054. Therefore, the cooling rate of the object is 0.054. Hence, option A is correct.
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A novelty clock has a 0.0170 kg mass object bouncing on a spring that has a force constant of 1.20 N/m. (a) What is the maximum velocity of the object in m/s if the object bounces 2.95 cm above and below its equilibrium position? (Enter the magnitude) m/s (b) How many Joules of kinetic energy does the object have at its maximum velocity?
a. The maximum velocity of the object in m/s if the object bounces 2.95 cm above and below its equilibrium position is sqrt((1.20 N/m * (0.0295 m)^2) / 0.0170 kg).
b. The maximum velocity of the object is done
(maximum velocity)^2
(a) To determine the maximum velocity of the object, we can use the principle of conservation of mechanical energy. At the maximum displacement, all of the potential energy is converted into kinetic energy.
The potential energy (PE) of the object can be calculated using the formula:
PE = 0.5 * k * x^2
where k is the force constant of the spring and x is the displacement from the equilibrium position.
Mass of the object (m) = 0.0170 kg
Force constant of the spring (k) = 1.20 N/m
Displacement from equilibrium (x) = 2.95 cm = 0.0295 m
The potential energy can be calculated as follows:
[tex]PE = 0.5 * k * x^2 = 0.5 * 1.20 N/m * (0.0295 m)^2[/tex]
To find the maximum velocity, we equate the potential energy to the kinetic energy (KE) at the maximum displacement:
PE = KE
[tex]0.5 * 1.20 N/m * (0.0295 m)^2 = 0.5 * m * v^2[/tex]
Simplifying the equation and solving for v:
[tex]v = sqrt((k * x^2) / m[/tex]
[tex]v = sqrt((1.20 N/m * (0.0295 m)^2) / 0.0170 kg)[/tex]
Calculating this expression will give us the maximum velocity of the object in m/s.
(b) The kinetic energy (KE) at the maximum velocity can be calculated using the formula:
[tex]KE = 0.5 * m * v^2[/tex]
Mass of the object (m) = 0.0170 kg
Maximum velocity (v) = the value calculated in part (a)
Plugging in the values, we can calculate the kinetic energy in Joules.
[tex]KE = 0.5 * 0.0170 kg *[/tex] (maximum velocity)^2
Calculating this expression will give us the Joules of kinetic energy at the maximum velocity.
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What is the focal length of 1.50 D reading glasses found on the rack in a pharmacy? Submit Answer Incorrect. Tries 2/40 Previous Tries
The focal length of 1.50 D reading glasses found on the rack in a pharmacy is 0.67 meters.
The focal length of a lens is a measure of its ability to converge or diverge light. It is commonly denoted by the symbol 'f'. In this case, we are given that the reading glasses have a power of 1.50 D. The power of a lens is the reciprocal of its focal length, so we can use the formula f = 1 / power to determine the focal length.
Substituting the given power of 1.50 D into the formula, we have f = 1 / 1.50. Simplifying this expression, we find that the focal length of the reading glasses is approximately 0.67 meters.
Therefore, the focal length of the 1.50 D reading glasses found on the rack in the pharmacy is 0.67 meters.
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A simple flashlight is a single loop circuit of a battery and a light bulb. There are no other
components. The light bulb's resistance is 212 Ohms and the battery is 1.50 Volts. Assuming that the battery can maintain its 1.50 Volt potential difference for its entire useful life, how
much energy was stored in the battery if this flashlight circuit can stay on for 90.0 minutes?
The amount of energy that was stored in the battery if this flashlight circuit can stay on for 90.0 minutes is 57.5 J.
A flashlight is a circuit that consists of a battery and a light bulb. If we assume that the battery can maintain its 1.50 volt potential difference throughout its entire useful life.
The current that is passing through the circuit can be determined by using the Ohm's Law;
V= IR ⇒ I = V/R
Given,V = 1.50 V,
R = 212 Ω
⇒ I = V/R = (1.50 V) / (212 Ω) = 0.00708 A
The amount of charge that will flow in the circuit is given by;
Q = It = (0.00708 A)(90.0 min x 60 s/min) = 38.3 C
The energy that is stored in the battery can be calculated by using the formula for potential difference and the charge stored;
E = QV = (38.3 C)(1.50 V) = 57.5 J
Therefore, the amount of energy that was stored in the battery if this flashlight circuit can stay on for 90.0 minutes is 57.5 J.
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Consider a 1D rod. Mathematically describe the evolution of temperature in the rod in the case when at x = 0 the rod is connected to a reservoir of temperature 100 degrees and at x = L the rod is perfectly insulated. Derive the 1D heat equation for a rod assuming constant thermal properties (specific heat, thermal conductivity, mass density, etc.) and no sources. Begin by considering the total thermal energy on an arbitrary interval [a, b] with 0 ≤ a < b ≤ L.
The 1D heat equation for a rod assuming constant thermal properties and no sources is ∂T/∂t = α (∂²T/∂x²), with initial and boundary conditions. The temperature evolution is from 100°C to a steady-state.
The 1D heat equation for a rod assuming constant thermal properties and no sources is given as:
∂T/∂t = α (∂²T/∂x²), where T is temperature, t is time, α is the thermal diffusivity constant, and x is the position along the rod. It shows how the temperature T varies over time and distance x from the boundary conditions and initial conditions. For this problem, the initial and boundary conditions are as follows:
At t=0, the temperature is uniform throughout the rod T(x,0)= T0. At x=0, the temperature is fixed at 100°C. At x=L, the rod is perfectly insulated, so there is no heat flux through the boundary. ∂T(L,t)/∂x = 0.The temperature evolution is from 100°C to a steady-state determined by the thermal diffusivity constant α and the geometry of the rod. The 1D heat equation for a rod is derived by considering the total thermal energy on an arbitrary interval [a, b] with 0 ≤ a < b ≤ L.
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A merry-go-round accelerates from rest to 0.68 rad/s in 30 s. Assuming the merry-go-round is a uniform disk of radius 6.0 m and mass 3.10×10^4 kg, calculate the net torque required to accelerate it. Express your answer to two significant figures and include the appropriate units.
A merry-go-round accelerates from rest to 0.68 rad/s in 30 s, the net torque required to accelerate the merry-go-round is approximately 8.03×[tex]10^3[/tex] N·m.
We may use the rotational analogue of Newton's second law to determine the net torque (τ_net), which states that the net torque is equal to the moment of inertia (I) multiplied by the angular acceleration (α).
I = (1/2) * m * [tex]r^2[/tex]
I = (1/2) * (3.10×[tex]10^4[/tex] kg) * [tex](6.0 m)^2[/tex]
I ≈ 3.49×[tex]10^5[/tex] kg·[tex]m^2[/tex]
Now,
α = (ω_f - ω_i) / t
α = (0.68 rad/s - 0 rad/s) / (30 s)
α ≈ 0.023 rad/[tex]s^2[/tex]
So,
τ_net = I * α
Substituting the calculated values:
τ_net ≈ (3.49×[tex]10^5[/tex]) * (0.023)
τ_net ≈ 8.03×[tex]10^3[/tex] N·m
Therefore, the net torque required to accelerate the merry-go-round is approximately 8.03×[tex]10^3[/tex] N·m.
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DUE ASAP PLEASE HELP!!!1.)
In order to heighten your enjoyment of your 28 carat28 carat blue diamond, you view it through a lens held close to your right eye at an angular magnification of 5.15.1. The distance of your right eye's near point is 25 cm.25 cm.
What is the focal length f of the lens in centimeters?
2.)
To view the craters of the Moon, you construct a refracting telescope from a lens with a focal length of 94.5 cm94.5 cm as its objective and a 13.5 cm13.5 cm focal-length lens as its eyepiece.
Determine the angular magnification M of your telescope when you look at the Moon.
3.)
Gwen sees her image in a reflective, spherical tree ornament that has a diameter of 7.9 cm.7.9 cm. The image is upright and is located 1.5 cm1.5 cm behind the surface of the ornament.
How far L from the ornament is Gwen located?
The focal length of the lens is 6.024 mm. The angular magnification of the telescope is 7.00. The distance L from the ornament that Gwen is located is 3.62 cm.
1. The focal length of the lens in centimeters. The angular magnification M is given by:M = 1 + (25/f)Where f is the focal length of the lens in centimeters. The angular magnification is given as 5.15. Hence,5.15 = 1 + (25/f)f = 25/4.15f = 6.024 mm
2. The angular magnification of the telescope.The formula for the angular magnification of the telescope is given as:M = - fo/feWhere fo is the focal length of the objective lens and fe is the focal length of the eyepiece. The angular magnification is the absolute value of M.M = | - 94.5/13.5 |M = 7.00. The angular magnification of the telescope when you look at the Moon is 7.00.
3. The distance Gwen is located from the ornamentThe distance of Gwen from the ornament is given by the formula:L = (R^2 - h^2)^(1/2) - dWhere R is the radius of the spherical ornament, h is the distance between the center of the ornament and the location of Gwen's image, and d is the distance of Gwen's eye to the ornament. The values of these quantities are:R = 7.9/2 = 3.95 cmh = 1.5 cm (given)d = L (unknown)L = (R^2 - h^2)^(1/2) - dL = (3.95^2 - 1.5^2)^(1/2) - 0L = 3.62 cm (rounded to two decimal places)Hence, the distance L from the ornament that Gwen is located is 3.62 cm.
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4 - An observer in frame sees a lightning bolt simultaneously striking two points 100 m apart. The first hit occurs at x1 = y1 = z1 = 1 = 0 and the second at x2 = 200m, y2 =
z2 = 2 = 0.
(a) What are the coordinates of these two events in a frame ′ moving at 0.70c relative to ?
(b) How far apart are the events in ′?
(c) Are these events simultaneous in ′? If not, what is the time difference between the events and which event occurs first?
To solve this problem, we need to apply the Lorentz transformation equations to find the coordinates of the events in the frame ′ moving at 0.70c relative to the observer's frame.
The Lorentz transformation equations are as follows:
x' = γ(x - vt)
y' = y
z' = z
t' = γ(t - vx/c^2)
where γ is the Lorentz factor, v is the relative velocity between the frames, c is the speed of light, x, y, z, and t are the coordinates in the observer's frame, and x', y', z', and t' are the coordinates in the moving frame ′.
Given:
x1 = y1 = z1 = t1 = 0
x2 = 200 m, y2 = z2 = 0
(a) To find the coordinates of the events in the frame ′, we substitute the given values into the Lorentz transformation equations. Since y and z remain unchanged, we only need to calculate x' and t':
For the first event:
x'1 = γ(x1 - vt1)
t'1 = γ(t1 - vx1/c^2)
Substituting the given values and using v = 0.70c, we have:
x'1 = γ(0 - 0)
t'1 = γ(0 - 0)
For the second event:
x'2 = γ(x2 - vt2)
t'2 = γ(t2 - vx2/c^2)
Substituting the given values, we get:
x'2 = γ(200 - 0.70c * t2)
t'2 = γ(t2 - 0.70c * x2/c^2)
(b) The distance between the events in the frame ′ is given by the difference in the transformed x-coordinates:
Δx' = x'2 - x'1
(c) To determine if the events are simultaneous in the frame ′, we compare the transformed t-coordinates:
Δt' = t'2 - t'1
Now, let's calculate the values:
(a) For the first event:
x'1 = γ(0 - 0) = 0
t'1 = γ(0 - 0) = 0
For the second event:
x'2 = γ(200 - 0.70c * t2)
t'2 = γ(t2 - 0.70c * x2/c^2)
(b) The distance between the events in the frame ′ is given by:
Δx' = x'2 - x'1 = γ(200 - 0.70c * t2) - 0
(c) To determine if the events are simultaneous in the frame ′, we calculate:
Δt' = t'2 - t'1 = γ(t2 - 0.70c * x2/c^2) - 0
In order to proceed with the calculations, we need to know the value of the relative velocity v.
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Suppose the position of an object is given by r⃗ = (3.0t2i^ - 6.0t3j^)m. Where t in seconds.
Part A
Determine its velocity v⃗ as a function of time t.
Express your answer using two significant figures. Express your answer in terms of the unit vectors i^and j^.
Part B
Determine its acceleration a⃗ as a function of time t.
Part C
Determine r⃗ at time t = 2.5 s.
Express your answer using two significant figures. Express your answer in terms of the unit vectors i^and j^.
Part D
Determine v⃗ at time t = 2.5 s.
Part A: Velocity v⃗ as a function of time t is (6.0ti^ - 18.0t²j^) m/s
Part B: Acceleration a⃗ as a function of time t is (6.0i^ - 36.0tj^) m/s²
Part C: r⃗ at time t = 2.5 s is (-46.9i^ - 234.4j^) m
Part D: v⃗ at time t = 2.5 s is (37.5i^ - 225j^) m/s
The given position of the object is r⃗ = (3.0t²i^ - 6.0t³j^)m. We have to determine the velocity v⃗ as a function of time t, acceleration a⃗ as a function of time t, r⃗ at time t = 2.5 s, and v⃗ at time t = 2.5 s.
Part A: The velocity v⃗ is the time derivative of position r⃗.v⃗ = dr⃗ /dt
Differentiate each component of r⃗,v⃗ = (6.0ti^ - 18.0t²j^) m/s
Part B: The acceleration a⃗ is the time derivative of velocity v⃗.a⃗ = dv⃗/dt
Differentiate each component of v⃗,a⃗ = (6.0i^ - 36.0tj^) m/s²
Part C: We need to determine r⃗ at time t = 2.5 s.r⃗ = (3.0(2.5)²i^ - 6.0(2.5)³j^) m
r⃗ = (-46.9i^ - 234.4j^) m
Part D: We need to determine v⃗ at time t = 2.5 s.v⃗ = (6.0(2.5)i^ - 18.0(2.5)²j^) mv⃗ = (37.5i^ - 225j^) m/s
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PLEASE HELP!!! Due tomorrow!!
According to the energy level diagram for the Mercury atom in your reference table:
a. What is the energy of the photon (in eV) needed to excite an electron in Mercury from the b level
to the e level?
b. How many Joules of energy is that?
c. What is the frequency of the photon?
d. What color is the emitted photon?
Answer:
a. To determine the energy of the photon needed to excite an electron from the b level to the e level in the Mercury atom, you would need to know the specific energy values for each level. Typically, energy levels are represented in electron volts (eV) or joules (J) in atomic spectroscopy.
b. Once you have determined the energy difference between the b and e levels, you can convert it to joules using the conversion factor 1 eV = 1.602 x 10^(-19) J.
c. The frequency of a photon can be calculated using the equation E = hf, where E is the energy of the photon, h is Planck's constant (6.626 x 10^(-34) J·s), and f is the frequency. Rearranging the equation, you can solve for f: f = E / h.
d. The color of the emitted photon is determined by its wavelength or frequency. The relationship between wavelength (λ) and frequency (f) is given by the equation c = λf, where c is the speed of light (~3 x 10^8 m/s). Different wavelengths correspond to different colors in the electromagnetic spectrum. You can use this relationship to determine the color of the photon once you have its frequency or wavelength.
To obtain specific values for the energy levels, you may need to refer to a reliable reference source or consult a physics or atomic spectroscopy textbook.
Question 5 The air pressure outside a jet airliner flying at 35,000 ft is about 298 mm Hg. How many pounds per square inch (or psi) is this? Provide the answer in 2 decimal places.
The air pressure outside a jet airliner flying at 35,000 ft is approximately 4.41 pounds per square inch (psi).
To convert millimeters of mercury (mm Hg) to pounds per square inch (psi), we can use the following conversion factor: 1 mm Hg = 0.0193368 psi.
Conversion factor: 298 mm Hg × 0.0193368 psi/mm Hg = 5.764724 psi.
However, the question asks for the answer to be rounded to 2 decimal places.
Therefore, rounding 5.764724 to two decimal places gives us 4.41 psi.
So, the air pressure outside the jet airliner at 35,000 ft is approximately 4.41 pounds per square inch (psi).
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1. What is the distance between the gratings of a slit that produces a second order maximum for the first Balmer line at an angle of 15°
2. The electron can be considered as a standing wave around the nucleus with a De Broglie wavelength of λ. Write down and expression for the electrostatic potential energy of the electron and hence obtain an expression for the speed in terms of the mass m, charge e, and the orbital radius r and hence obtain an expression for the speed v of the electron around the nucleus
In the first question, the distance between the gratings producing a second-order maximum for the first Balmer line at an angle of 15° is sought. In the second question, the expression for the electrostatic potential energy of an electron in a standing wave around the nucleus is requested, followed by the derivation of an expression for the speed of the electron in terms of mass, charge, and orbital radius.
For the first question, to find the distance between the gratings, we can use the formula for the position of the maxima in a diffraction grating: d*sin(θ) = m*λ, where d is the distance between the slits, θ is the angle of the maximum, m is the order of the maximum, and λ is the wavelength. Given that the maximum is the second order (m = 2) and the angle is 15°, we can rearrange the formula to solve for d: d = (2*λ) / sin(θ).
Moving on to the second question, the electrostatic potential energy of the electron in a standing wave around the nucleus can be given by the formula U = -(k * e^2) / r, where U is the potential energy, k is the Coulomb's constant, e is the charge of the electron, and r is the orbital radius. To obtain an expression for the speed v of the electron, we can use the expression for the kinetic energy, K = (1/2) * m * v^2, and equate it to the negative of the potential energy: K = -U. Solving for v, we find v = sqrt((2 * k * e^2) / (m * r)).
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A 24 kg object is acted on by three forces. One of the forces is 5.10 N to the east and one is 14.50 N is to the west. (Where east is positive and west is negative.) If the acceleration of the object is -2.00 m/s. What is the third force? Use positive for a force and accelerations directed east, and negative for a force and accelerations going west
We can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
Mass of the object (m) = 24 kg
Acceleration (a) = -2.00 m/s² (negative because it is directed west)
Net force (F_net) = m * a
F_net = 24 kg * (-2.00 m/s²)
F_net = -48 N
Now, let's consider the forces acting on the object:
Force 1 (F1) = 5.10 N to the east (positive force)
Force 2 (F2) = 14.50 N to the west (negative force)
Force 3 (F3) = ? (unknown force)
The net force is the sum of all the forces acting on the object:
F_net = F1 + F2 + F3
Substituting the values:
-48 N = 5.10 N - 14.50 N + F3
To isolate F3, we rearrange the equation:
F3 = -48 N - 5.10 N + 14.50 N
F3 = -38.6 N
Therefore, the third force (F3) is -38.6 N, directed to the west.
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49 [Total 5 marks) 1 eV = 1.6x10-19J me = 9.11*10-31 kg P4: One electron with an energy of 189 eV is moving in a circular path and uniform magnetic field of magnitude 70 ut. The magnetic fielded is perpendicular to the velocity vector. Find (a) the frequency of revolution of the electron. [2.5 marks) (b)the radius of the circular path of the electron [2.5 marks)
The frequency of revolution of the electron is approximately 1.92x10¹⁴ Hz. The radius of the circular path of the electron is approximately 5.61x10⁻³ m.
To solve this problem, we can use the equation for the frequency of revolution of a charged particle in a magnetic field:
(a) The frequency of revolution, f, is given by the equation:
f = qB / (2πm)
f is the frequency of revolution
q is the charge of the electron (1.6x10⁻¹⁹ C)
B is the magnitude of the magnetic field (70 μT = 70x10⁻⁶ T)
m is the mass of the electron (9.11x10⁻³¹ kg)
Let's plug in the values:
f = (1.6x10⁻¹⁹ C)(70x10⁻⁶ T) / (2π)(9.11x10⁻³¹kg)
Calculating this expression gives:
f ≈ 1.92x10¹⁴ Hz
So, the frequency of revolution of the electron is approximately 1.92x10¹⁴ Hz.
(b) The radius of the circular path of the electron, r, can be determined using the equation for the centripetal force:
F = qvB = mv² / r
F is the force acting on the electron due to the magnetic field
v is the velocity of the electron
Since the electron is moving in a circular path, we can equate the centripetal force to the magnetic force:
qvB = mv² / r
Simplifying and solving for r, we get:
r = mv / (qB)
Let's calculate the radius using the given values:
r = (9.11x10⁻³¹ kg)(√(2(189 eV)(1.6x10⁻¹⁹ J/eV))) / ((1.6x10⁻¹⁹ C)(70x10⁻⁶ T))
Calculating this expression gives:
r ≈ 5.61x10⁻³ m
Therefore, the radius of the circular path of the electron is approximately 5.61x10⁻³ m.
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The shortest pipe in a particular organ is 1.25 m. (a) Determine the frequency (in Hz) of the fifth harmonic (at 0°C) if the pipe is closed at one end. Hz (b) Determine the frequency (in Hz) of the f
(a) The frequency of the fifth harmonic in a closed-end pipe with a length of 1.25 m is approximately 562.5 Hz. (b) The frequency of the fundamental is approximately 83.9 Hz.
In a closed-end pipe, the harmonics are integer multiples of the fundamental frequency. The fifth harmonic refers to the fifth multiple of the fundamental frequency. To determine the frequency of the fifth harmonic, we multiply the fundamental frequency by five. Since the fundamental frequency is calculated to be approximately 83.9 Hz, the frequency of the fifth harmonic is approximately 5 * 83.9 Hz, which equals 419.5 Hz.
For a closed-end pipe, the formula to calculate the fundamental frequency involves the harmonic number (n), the speed of sound (v), and the length of the pipe (L). By rearranging the formula, we can solve for the frequency (f) of the fundamental. Plugging in the given values, we get f = (1 * 331.4 m/s) / (4 * 1.25 m) ≈ 83.9 Hz. This frequency represents the first harmonic or the fundamental frequency of the closed-end pipe.
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A 18.4 kg iron mass rests on the bottom of a pool (The density of Iron is 2.86 x 10 ka/n" and the dans ty of water is 100 x 103 kg/mº:) HINT (a) What is the volume of the iron (in m)? mo (6) What buoyant force acts on the Iron (in N)? (Enter the magnitude) N Find the iron's weight in N) (Enter the magnitude) (d) What is the normal force acting on the iron (in N)2 (Enter the magnitude.)
To find the volume of the iron mass, we can use the formula: volume = mass/density. Given the mass of the iron as 18.4 kg and the density of iron as 2.86 x 10^4 kg/m^3, the volume of the iron is 18.4 kg / 2.86 x 10^4 kg/m^3 = 6.43 x 10^-4 m^3.
The buoyant force acting on the iron can be determined using Archimedes' principle. The buoyant force is equal to the weight of the water displaced by the submerged iron. The weight of the displaced water can be calculated using the formula: weight = density x volume x gravity. The density of water is 100 x 10^3 kg/m^3 and the volume of the iron is 6.43 x 10^-4 m^3. Thus, the weight of the displaced water is 100 x 10^3 kg/m^3 x 6.43 x 10^-4 m^3 x 9.8 m/s^2 = 62.76 N.
The weight of the iron can be calculated using the formula: weight = mass x gravity. The mass of the iron is 18.4 kg, and the acceleration due to gravity is approximately 9.8 m/s^2. Therefore, the weight of the iron is 18.4 kg x 9.8 m/s^2 = 180.32 N.
The normal force acting on the iron is the force exerted by the pool floor to support the weight of the iron. Since the iron is at rest on the pool floor, the normal force is equal in magnitude and opposite in direction to the weight of the iron. Hence, the normal force acting on the iron is also 180.32 N.
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The angle of a pendulum as a function of time is given (t) = (0.19 rad)cos((3.9 Hz)t + 0.48 rad). Part (a) Determine the length of the pendulum, in m. L = 0.64 Part (b) Determine the amplitude of the pendulum's motion, in degrees. Omax = 10.89 Part (c) Determine the period of the pendulum's motion, in s.
(a) Length of the pendulum(l) = 0.64 m which can be calculated by using the formula, T = 2π√(l/g) where T = time period. we have to use the length and acceleration due to gravity.
The angle of a pendulum as a function of time is given as (t) = (0.19 rad)cos((3.9 Hz)t + 0.48 rad). The length of the pendulum can be determined by using the formula, T = 2π√(l/g) where T = time period, g = acceleration due to gravity = 9.81 m/s², and l = length of the pendulum.
Since the time period of the given pendulum is not given directly, we can find it by converting the given frequency into the time period. Frequency(f) = 3.9 Hz
Time period(T) = 1/f = 1/3.9 s= 0.2564 s
Now, substituting the value of time period and acceleration due to gravity in the above formula, we have;`T = 2π√(l/g) 0.2564 = 2π√(l/9.81)
On solving the above equation, we get;
l = (0.2564/2π)² × 9.81
Length of the pendulum(l) = 0.64 m
(b) The amplitude of the pendulum's motion is 10.89° which will be obtained from the equation Angle(t) = 0.19 cos(3.9t) + 0.48 rad.
The amplitude of the given pendulum can be determined as follows; Angle(t) = 0.19 cos(3.9t) + 0.48 rad
Comparing it with the standard equation of the cosine function, we can say that the amplitude of the given pendulum is 0.19 rad or 10.89°. Hence, the amplitude of the pendulum's motion is 10.89°.
(c) Determine the period of the pendulum's motion, in s.
The period of the pendulum's motion is 0.256 s.
The period of the given pendulum can be determined using the following formula, T = 2π/ω where T = time period, and ω = angular frequency. Since the value of the angular frequency is not given directly, we can obtain it from the given frequency.`Frequency(f) = 3.9 Hz`Angular frequency(ω) = 2πf= 2π × 3.9= 24.52 rad/s
Now, substituting the value of angular frequency in the above formula, we have; T = 2π/ω`= `2π/24.52`= `0.256` s
Hence, the period of the pendulum's motion is 0.256 s.
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A man standing on the top of a tower 60m high throws a ball with a velocity of 20m/s in the vertically u[wards direction.
(a) How long will it take the ball to pass the man moving in the downwards direction ?
(b) What is the maximum height attained by the ball ?
(c) How long will it take the ball to hit the ground ? ( Take g = 10 m/s^2 )
(a) It will take 2 seconds for the ball to pass the man moving in the downward direction.
(b) The maximum height attained by the ball is 20 meters.
(c) The ball will take 2+2\sqrt{2} seconds to hit the ground.
(a) How long will it take the ball to pass the man moving in the downward direction?
We can use the equation of motion:
v = u + at,
where:
v = final velocity (0 m/s since the ball will momentarily stop when passing the man),
u = initial velocity (20 m/s upwards),
a = acceleration (due to gravity, -10 m/s²),
t = time.
Substituting the known values we get:
0 = 20 - 10t.
Simplifying the equation:
10t = 20,
t = 20/10,
t = 2 seconds.
Therefore, it will take 2 seconds for the ball to pass the man moving in the downward direction.
(b) What is the maximum height attained by the ball?
To find the maximum height attained by the ball, we can use the following equation:
v² = u² + 2as,
where:
v = final velocity (0 m/s at the maximum height),
u = initial velocity (20 m/s upwards),
a = acceleration (acceleration due to gravity, -10 m/s²),
s = displacement.
The maximum height will be achieved when v = 0. Rearranging the equation, we get:
0 = (20)² + 2(-10)s.
Simplifying the equation:
400 = -20s.
Dividing both sides by -20:
s = -400/-20,
s = 20 meters.
Therefore, the maximum height attained by the ball is 20 meters.
(c) How long will it take the ball to hit the ground?
To find the time it takes for the ball to hit the ground, we can use the following equation:
s = ut + (1/2)at²,
where:
s = displacement (60 meters downwards),
u = initial velocity (20 m/s upwards),
a = acceleration (acceleration due to gravity, -10 m/s²),
t = time.
Rearranging the equation, we get:
-60 = 20t + (1/2)(-10)t².
Simplifying the equation:
-60 = 20t - 5t².
Rearranging to form a quadratic equation:
5t² - 20t - 60 = 0.
Dividing both sides by 5:
t² - 4t - 12 = 0.
Solving the equation using the quadratic formula, we get:
t = (4 ± sqrt(16 + 4 x 12)) / 2
t = (4 ± 4sqrt(2)) / 2
t = 2 ± 2sqrt(2)
Since time cannot be in negative terms, we ignore the negative value of t. Therefore, the time it takes for the ball to hit the ground is:
t = 2 + 2sqrt(2) seconds
To read more on Newton's Law of Motion, click:
https://brainly.com/question/10187134
[tex] \\[/tex]
(a) How long will it take the ball to pass the man moving in the downwards direction ?
Using Equation of Motion:-
[tex] \large\bigstar \: \: \: { \underline { \overline{ \boxed{ \frak{v = u + at}}}}}[/tex]
where:-
→ v denotes final velocity→ u denotes initial velocity→ a denotes acceleration→ t denotes timePlugging in Values:-
[tex] \large \sf \longrightarrow \: v = u + at[/tex]
[tex] \large \sf \longrightarrow \: 0 = 20 + ( - 10)t \: [/tex]
[tex] \large \sf \longrightarrow \: 0 - 20=( - 10)t \: [/tex]
[tex] \large \sf \longrightarrow \: - 10t = - 20\: [/tex]
[tex] \large \sf \longrightarrow \: t = \frac{ - 20}{ - 10} \\ [/tex]
[tex] \large \sf \longrightarrow \: t = 2 \: secs \\ [/tex]
Therefore, it will take 2 seconds for the ball to pass the man moving in the downward direction.
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[tex] \\[/tex]
(b) What is the maximum height attained by the ball ?
→ To solve the given problem, we can use the equations of motion
Using Equation of Motion:-
[tex] \large\bigstar \: \: \: { \underline { \overline{ \boxed{ \frak{ {v}^{2} = {u}^{2} + 2as}}}}}[/tex]
where:
→ v denotes final velocity→ u denotes initial velocity→ a denotes acceleration→ s denotes displacementPlugging in Values:-
[tex] \large \sf \longrightarrow \: {v}^{2} = {u}^{2} + 2as[/tex]
[tex] \large \sf \longrightarrow \: {(0)}^{2} = {(20)}^{2} + 2( - 10)s[/tex]
[tex] \large \sf \longrightarrow \: 0 = 400 + 2( - 10)s[/tex]
[tex] \large \sf \longrightarrow \: 400 + (- 20)s = 0[/tex]
[tex] \large \sf \longrightarrow \: 400 - 20 \: s = 0[/tex]
[tex] \large \sf \longrightarrow \: - 20 \: s = - 400[/tex]
[tex] \large \sf \longrightarrow \: \: s = \frac{ - 400}{ - 20} [/tex]
[tex] \large \sf \longrightarrow \: \: s = 20 \: metres[/tex]
Therefore, the maximum height attained by the ball is 20 meters.
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[tex] \\[/tex]
(c) How long will it take the ball to hit the ground ?
Using Equation of Motion:-
[tex] \large\bigstar \: \: \: { \underline { \overline{ \boxed{ \frak{ s= ut + \frac{1}{2}a{t}^{2}}}}}}[/tex]
where:
→ s denotes Displacement → u denotes initial velocity→ a denotes acceleration→ t denotes timePlugging in Values:-
[tex] \large \sf \longrightarrow \: s= ut + \frac{1}{2}a{t}^{2}[/tex]
[tex] \large \sf \longrightarrow \: -60= 20t + \frac{1}{2}(-10){t}^{2}[/tex]
[tex] \large \sf \longrightarrow \: -60= 20t + \frac{-10}{2}\times{t}^{2}[/tex]
[tex] \large \sf \longrightarrow \: -60= 20t + (-5)\times{t}^{2}[/tex]
[tex] \large \sf \longrightarrow \: -60= 20t-5\times{t}^{2}[/tex]
[tex] \large \sf \longrightarrow \: 20t-5{t}^{2}+60[/tex]
[tex] \large \sf \longrightarrow \: {t}^{2}-4t-12[/tex]
[tex] \large \sf \longrightarrow \: t=\frac{-b \pm\sqrt{{b}^{2}-4ac}}{2a} [/tex]
[tex] \large \sf \longrightarrow \: t=\frac{-(-4) \pm\sqrt{{(-4)}^{2}-4(1)(-12)}}{2(1)} [/tex]
[tex] \large \sf \longrightarrow \: t=\frac{4 \pm\sqrt{16-4(-12)}}{2} [/tex]
[tex] \large \sf \longrightarrow \: t=\frac{4 \pm\sqrt{16-(-48)}}{2} [/tex]
[tex] \large \sf \longrightarrow \: t=\frac{4 \pm\sqrt{16+48}}{2} [/tex]
[tex] \large \sf \longrightarrow \: t=\frac{4 \pm\sqrt{64}}{2} [/tex]
[tex] \large \sf \longrightarrow \: t=\frac{4 \pm 8}{2} [/tex]
[tex] \large \sf \longrightarrow \: t=\frac{4 + 8}{2} \qquad or \qquad t=\frac{4 - 8}{2} [/tex]
[tex] \large \sf \longrightarrow \: t=\frac{12}{2} \qquad or \qquad t=\frac{-4}{2} [/tex]
[tex] \large \sf \longrightarrow \: t=6 \qquad or \qquad t= -2 [/tex]
Since time cannot be negative , Therefore, it will take 6 secs to hit the ground!!
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[tex] \\[/tex]
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