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

Answers

Answer 1

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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Related Questions

A 11.9 g bullet traveling at unknown speed is fired into a 0.317 kg wooden block anchored to a 120 N/m spring. What is the speed of the bullet (in m/sec) if the spring is compressed by 43.5 cm before the combined block/bullet comes to stop?

Answers

The speed of the bullet is approximately 156.9 m/s.

To find the speed of the bullet, we need to consider the conservation of momentum and energy in the system.

Let's assume the initial speed of the bullet is v. The mass of the bullet is given as 11.9 g, which is equal to 0.0119 kg. The wooden block has a mass of 0.317 kg.

According to the conservation of momentum, the momentum before the collision is equal to the momentum after the collision. The momentum of the bullet is given by its mass multiplied by its initial velocity, while the momentum of the combined block and bullet system after the collision is zero since it comes to a stop.

So, we have:

(m_bullet)(v) = (m_block + m_bullet)(0)

(0.0119 kg)(v) = (0.0119 kg + 0.317 kg)(0)

This equation tells us that the velocity of the bullet before the collision is 0 m/s. However, this does not make sense physically since the bullet was fired into the wooden block.

Therefore, there must be another factor at play: the compression of the spring. When the bullet collides with the wooden block, their combined energy is transferred to the spring, causing it to compress.

We can calculate the potential energy stored in the compressed spring using Hooke's Law:

Potential energy = (1/2)kx^2

where k is the spring constant and x is the compression of the spring. In this case, the spring constant is given as 120 N/m, and the compression is 43.5 cm, which is equal to 0.435 m.

Potential energy = (1/2)(120 N/m)(0.435 m)^2

Next, we equate this potential energy to the initial kinetic energy of the bullet:

Potential energy = (1/2)m_bullet*v^2

(1/2)(120 N/m)(0.435 m)^2 = (1/2)(0.0119 kg)(v)^2

Simplifying the equation, we can solve for v:

(120 N/m)(0.435 m)^2 = (0.0119 kg)(v)^2

v^2 = [(120 N/m)(0.435 m)^2] / (0.0119 kg)

Taking the square root of both sides, we get:

v ≈ 156.9 m/s

Therefore, the speed of the bullet is approximately 156.9 m/s.

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50. The angle that a reflected light ray makes with the surface normal A) is smaller B) the same size C) greater than the angle that the incident ray makes with the normal 51. The speed of light in gl

Answers

The angle that a reflected light ray makes with the surface normal is smaller.

The law of reflection states that the angle of incidence is equal to the angle of reflection. When light is reflected from a surface, the angle at which it is reflected (angle of reflection) is equal to the angle at which it hits the surface (angle of incidence). The angle that a reflected light ray makes with the surface normal is the angle of reflection. Therefore, the answer is that the angle that a reflected light ray makes with the surface normal is smaller than the angle that the incident ray makes with the normal.

The speed of light in glass is less than the speed of light in a vacuum. This means that the refractive index of glass is greater than 1. When light passes through a medium with a higher refractive index than the medium it was previously in, the light is bent towards the normal. Therefore, the answer is that the speed of light in glass is less than the speed of light in a vacuum, and the refractive index of glass is greater than 1.

The angle that a reflected light ray makes with the surface normal is A) is smaller. The law of reflection states that the angle of incidence is equal to the angle of reflection. When light is reflected from a surface, the angle at which it is reflected (angle of reflection) is equal to the angle at which it hits the surface (angle of incidence). The angle that a reflected light ray makes with the surface normal is the angle of reflection. Therefore, the answer is that the angle that a reflected light ray makes with the surface normal is smaller than the angle that the incident ray makes with the normal.

The speed of light in glass is less than the speed of light in a vacuum. This means that the refractive index of glass is greater than 1. When light passes through a medium with a higher refractive index than the medium it was previously in, the light is bent towards the normal. Therefore, the answer is that the speed of light in glass is less than the speed of light in vacuum, and the refractive index of glass is greater than 1.


When a light wave strikes a surface, it can be either absorbed or reflected. Reflection occurs when light bounces back from a surface. The angle at which the light strikes the surface is known as the angle of incidence, and the angle at which it reflects is known as the angle of reflection. The angle of incidence is always equal to the angle of reflection, as stated by the law of reflection. The angle that a reflected light ray makes with the surface normal is the angle of reflection. It's smaller than the angle of incidence.

When light travels through different mediums, such as air and glass, its speed changes, and it bends. Refraction is the process of bending that occurs when light moves from one medium to another with a different density. The refractive index is a measure of the extent to which a medium slows down light compared to its speed in a vacuum. The refractive index of a vacuum is 1.

When light moves from a medium with a low refractive index to a medium with a high refractive index, it bends toward the normal, which is a line perpendicular to the surface separating the two media.

When light is reflected from a surface, the angle of reflection is always equal to the angle of incidence. The angle of reflection is the angle that a reflected light ray makes with the surface normal, and it is smaller than the angle of incidence. The refractive index of a medium is a measure of how much the medium slows down light compared to its speed in a vacuum. When light moves from a medium with a low refractive index to a medium with a high refractive index, it bends toward the normal.

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JUNCTION RULE: (1) I 1
=I 3
+I 4
LOOP RULE: (2) LOOP I (LEFT CIRUT) V 0
−I 3
R 3
−I 3
R 2
−I 1
R 1
=0 LOOP 2 (RIGHT CIRCUT): (3) −I 4
R 4
+I 3
R 3
+I 3
R 3
=0

Answers

According to the junction rule, the current entering junction 1 is equal to the sum of the currents leaving junction 1: I1 = I3 + I4.

The junction rule, or Kirchhoff's current law, states that the total current flowing into a junction is equal to the total current flowing out of that junction. In this case, at junction 1, the current I1 is equal to the sum of the currents I3 and I4. This rule is based on the principle of charge conservation, where the total amount of charge entering a junction must be equal to the total amount of charge leaving the junction. Applying the loop rule, or Kirchhoff's voltage law, we can analyze the potential differences around the loops in the circuit. In the left circuit, traversing the loop in a clockwise direction, we encounter the potential differences V0, -I3R3, -I3R2, and -I1R1. According to the loop rule, the algebraic sum of these potential differences must be zero to satisfy the conservation of energy. This equation relates the currents I1 and I3 and the voltages across the resistors in the left circuit. Similarly, in the right circuit, traversing the loop in a clockwise direction, we encounter the potential differences -I4R4, I3R3, and I3R3. Again, the loop rule states that the sum of these potential differences must be zero, providing a relationship between the currents I3 and I4.

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A 83-ko pot in traing sites in a centuge that in his seat around a centras. When the setmaving in its chat a speed of 3.5 m/s, he feels a 455-N force bring against his back the seat faces the axis). What is the radius of the centrifuge 0.04 Xm

Answers

The radius of the centrifuge is 0.04 meters (m).

In this scenario, a person is seated in a centrifuge that rotates at a certain speed, causing them to experience a force against their back. We need to calculate the radius of the centrifuge based on the given information.

The force experienced by the person can be calculated using the formula for centripetal force:

Force = (Mass × Speed^2) / Radius

Given:

Force = 455 Newtons (N)

Speed = 3.5 meters per second (m/s)

Radius = 0.04 meters (m)

Plugging in the values into the formula, we can rearrange it to solve for the radius:

Radius = (Mass × Speed^2) / Force

Since the mass of the person (83 kg) is not given, we can solve for it by rearranging the formula:

Mass = (Force × Radius) / Speed^2

Mass = (455 N × 0.04 m) / (3.5 m/s)^2

Mass = (18.2 N·m) / 12.25 m^2/s^2

Mass ≈ 1.49 kg

Now that we have the mass, we can substitute it back into the formula for radius:

Radius = (Mass × Speed^2) / Force

Radius = (1.49 kg × (3.5 m/s)^2) / 455 N

Radius ≈ 0.04 m

The radius of the centrifuge is approximately 0.04 meters (m). This calculation is based on the given force experienced by the person (455 N) and the speed of the centrifuge (3.5 m/s). It assumes that the person's mass is 83 kilograms (kg). Please note that the accuracy of the result depends on the accuracy of the given values and assumptions made during the calculation.

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A negatively charged plastic bead is a distance d from the origin. At this moment, the magnitude of the electric held at the origin due to the bead is 369 N/C of the bead were moved so that it was a distance 3d from the origin, what would be the magnitude of the electric a field at the origin, in N/C?

Answers

A negatively charged plastic bead is a distance d from the origin. At this moment, the magnitude of the electric held at the origin due to the bead is 369 N/C of the bead were moved so that it was a distance 3d from the origin  if the plastic bead is moved to a distance 3d from the origin, the magnitude of the electric field at the origin would be 3321 N/C.

The magnitude of the electric field at a point due to a charged object can be calculated using the formula:

E = k × |Q| / r^2

where E is the electric field, k is the electrostatic constant (k = 9 × 10^9 N m^2/C^2), |Q| is the magnitude of the charge, and r is the distance from the charged object.

In the given scenario, the magnitude of the electric field at the origin (E1) due to the plastic bead at a distance d is 369 N/C.

We can use this information to determine the magnitude of the electric field at the origin (E2) if the bead is moved to a distance 3d from the origin.

Since the charge of the bead remains the same, the ratio of the electric fields is inversely proportional to the square of the distances:

E1 / E2 = (d^2) / (3d)^2

369 / E2 = 1 / 9

Solving for E2:

E2 = 9 ×369

E2 = 3321 N/C

Therefore, if the plastic bead is moved to a distance 3d from the origin, the magnitude of the electric field at the origin would be 3321 N/C.

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a) At an air show a jet flies directly toward the stands at a speed of 1180 km/h, emitting a frequency of 3810 Hz, on a day when the speed of sound is 342 m/s. What frequency in Ha) is received by the observers? Hz b) What frequency (in Hz) do they receive as the plane files directly away from them?

Answers

The observers perceive a frequency of around 3984.6 Hz when the jet flies directly toward them. As the plane flies directly away from the observers, they perceive a frequency of approximately 3655.4 Hz.

To calculate the frequency received by the observers, we need to consider the Doppler effect, which is the change in frequency of a wave due to the relative motion between the source and the observer.

f₀ = f ×  (v + v₀) / (v - vs)

where:

f₀ is the received frequency,

f is the emitted frequency,

v is the speed of sound,

v₀ is the velocity of the observer (0 in this case since they are stationary),

vs is the velocity of the source (1180 km/h converted to m/s).

Given:

f = 3810 Hz,

v = 342 m/s,

v₀= 0,

vs = 1180 km/h

   = (1180 × 1000) / 3600

    = 327.78 m/s

a) When the jet flies directly toward the stands, the observers perceive a higher frequency.

Plugging the values into the formula:

f₀= 3810 × (342 + 0) / (342 - 327.78)

f₀ ≈ 3984.6 Hz

Therefore, the observers receive a frequency of approximately 3984.6 Hz.

b) When the plane flies directly away from the observers, the perceived frequency is lower.

Given the same values as before:

f₀ = 3810 × (342 - 0) / (342 + 327.78)

f₀≈ 3655.4 Hz

Therefore, the observers receive a frequency of approximately 3655.4 Hz as the plane flies directly away from them.

Hence, the observers perceive a frequency of around 3984.6 Hz when the jet flies directly toward them. As the plane flies directly away from the observers, they perceive a frequency of approximately 3655.4 Hz.

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PORTUUR TEATER A ball is thrown directly downward with an initial speed of 8.25 m/s, from a height of 29.6 m. After what time interval does it strike the ground? Need Help? Read it Wasch PRACTILE ANUTHER

Answers

The ball will strike the ground after approximately 2.44 seconds, when the ball is thrown directly downward with an initial speed of 8.35 m/s.

Initial speed of the ball, u = 8.25 m/s

Height from which the ball is thrown, h = 29.6 m

We can use the kinematic equation of motion to find the time interval after which the ball will strike the ground.

The equation is given as v^2 = u^2 + 2gh

where v = final velocity of the ball = acceleration due to gravity = height from which the ball is thrown

We know that the ball will strike the ground when it will have zero vertical velocity. Thus, we can write the final velocity of the ball as 0.

Therefore, the above equation becomes:0 = u^2 + 2gh

Solving this equation for time, we get:t = sqrt(2h/g)

Substituting the given values, we get:

t = sqrt(2 × 29.6/9.81)≈ 2.44

Therefore, the ball will strike the ground after approximately 2.44 seconds.

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circular loop in the plane of the paper lies in a 0.63 T magnetic field pointing into the paper. If the loop's diameter changes from 20.0 cm to 8.0 cm in 0.71 s , what is the direction of the induced current? What is the magnitude of the average induced emf? Express your answer using two significant figures. If the coil resistance is 2.6 12 , what is the average induced current? Express your answer using two significant figures.

Answers

The direction of the induced current is counterclockwise when viewed from above the loop. The magnitude of the average induced emf is approximately 0.23 V. The direction of the induced current is opposite to the original current, and its magnitude is approximately 0.090 A.

To determine the direction of the induced current, we can apply Lenz's law, which states that the induced current creates a magnetic field that opposes the change in the magnetic flux through the loop.

Since the magnetic field points into the paper, the induced current will create a magnetic field that points out of the paper, opposing the original field. Therefore, the direction of the induced current is counterclockwise when viewed from above the loop.

Given that the loop's diameter changes from 20.0 cm to 8.0 cm in 0.71 s, we can calculate the average induced emf and the average induced current.

First, let's determine the change in magnetic flux (ΔΦ) through the loop. Since the loop lies in a magnetic field of 0.63 T, the magnetic field (B) remains constant.

The initial area (A_initial) of the loop can be calculated using the formula for the area of a circle: A_initial = π(r_initial)^2, where r_initial is the initial radius (half the initial diameter).

Similarly, the final area (A_final) of the loop is A_final = π(r_final)^2, where r_final is the final radius (half the final diameter).

The change in area (ΔA) is given by: ΔA = A_final - A_initial.

Let's plug in the values:

r_initial = 20.0 cm / 2 = 10.0 cm = 0.10 m

r_final = 8.0 cm / 2 = 4.0 cm = 0.04 m

A_initial = π(0.10 m)^2 = 0.0314 m²

A_final = π(0.04 m)^2 = 0.0050 m²

ΔA = A_final - A_initial = 0.0050 m² - 0.0314 m² = -0.0264 m² (negative due to decreasing area)

Now, we can calculate the average induced emf (ε_avg) using the formula:

ε_avg = -ΔΦ/Δt

where Δt is the time interval given as 0.71 s.

ε_avg = -(BΔA)/Δt = -(0.63 T)(-0.0264 m²)/(0.71 s) ≈ 0.234 V

The magnitude of the average induced emf is approximately 0.23 V (rounded to two significant figures).

Given that the coil resistance (R) is 2.6 Ω, we can now calculate the average induced current (I_avg) using Ohm's law:

I_avg = ε_avg / R

Substituting the values:

I_avg = 0.234 V / 2.6 Ω ≈ 0.090 A

The average induced current is approximately 0.090 A (rounded to two significant figures).

Therefore, the direction of the induced current is opposite to the original current, and its magnitude is approximately 0.090 A.

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Singly charged uranium-238 ions are accelerated through a potential difference of 2.00kV and enter a uniform magnetic field of magnitude 1.20 T directed perpendicular to their velocities.(c) What If? How does the ratio of these path radii depend on the accelerating voltage?

Answers

The ratio of the path radii for the uranium-238 ions is not affected by the accelerating voltage. The ratio is solely determined by the mass of the ions and the magnitude of the magnetic field.

The ratio of the path radii for singly charged uranium-238 ions depends on the accelerating voltage.

When a charged particle enters a uniform magnetic field perpendicular to its velocity, it experiences a force called the magnetic force. This force acts as a centripetal force, causing the particle to move in a circular path.

The magnitude of the magnetic force is given by the equation:
F = qvB
Where:

F is the magnetic force
q is the charge of the particle
v is the velocity of the particle
B is the magnitude of the magnetic field

In this case, the uranium-238 ions have a charge of +1 (since they are singly charged). The magnetic force acting on the ions is equal to the centripetal force:
qvB = mv²/r

Where:
m is the mass of the uranium-238 ion
v is the velocity of the ion
r is the radius of the circular path

We can rearrange this equation to solve for the radius:
r = mv/qB

The velocity of the ions can be determined using the equation for the kinetic energy of a charged particle:
KE = (1/2)mv²

The kinetic energy can also be expressed in terms of the accelerating voltage (V) and the charge (q) of the ion:
KE = qV

We can equate these two expressions for the kinetic energy:
(1/2)mv² = qV

Solving for v, we get:
v = sqrt(2qV/m)

Substituting this expression for v into the equation for the radius (r), we have:
r = m(sqrt(2qV/m))/qB

Simplifying, we get:
r = sqrt(2mV)/B

From this equation, we can see that the ratio of the path radii is independent of the charge (q) of the ions and the mass (m) of the ions.

Therefore, the ratio of the path radii is independent of the accelerating voltage (V).

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Four Small 0.200 Kg Spheres, Each Of Which You Can Regard As A Point Mass, Are Arranged In A Square 0.400 M On A Side And Connected By Light Rods.
Four small 0.200 kg spheres, each of which you can regard as a point mass, are arranged in a
square 0.400 m on a side and connected by light rods.
A 0.400 m 0.200 kg B (a) Find the moment of inertia of the system about an axis along the line CD. (b) The system starts to rotate from rest in the counterclockwise direction with an angular acceleration of + 2 rad/s². What is the angular velocity of the system after rotating 3 revolutions? (c) Calculate the rotational kinetic energy of the system. (KE-½Iw₂) (d) Calculate the angular momentum of the system. (L=Iw) (e) If the masses of spheres on the upper left and lower right were doubled, how would it affect your responses to (a) and (b) ?

Answers

(a) The moment of inertia of the system about an axis along the line CD is 0.038 kg·m².

(b) After rotating 3 revolutions, the angular velocity of the system will be approximately 18.85 rad/s.

(c) The rotational kinetic energy of the system is 0.717 J.

(d) The angular momentum of the system is 0.0754 kg·m²/s.

(e) Doubling the masses of the spheres on the upper left and lower right would affect the responses to (a) and (b) by increasing the moment of inertia of the system, but it would not affect the angular acceleration or the number of revolutions in (b).

(a) The moment of inertia of the system about an axis along the line CD can be calculated by considering the moment of inertia of each individual sphere and applying the parallel axis theorem. For a square arrangement, the moment of inertia of each sphere is 0.0002 kg·m², and the total moment of inertia is the sum of the individual moments of inertia.

(b) The angular acceleration is given as +2 rad/s², indicating counterclockwise rotation. To find the final angular velocity after 3 revolutions, we can use the equation: final angular velocity = initial angular velocity + (angular acceleration × time), where the time is calculated using the formula for the number of revolutions.

(c) The rotational kinetic energy of the system can be calculated using the formula KE = ½Iw², where I is the moment of inertia and w is the angular velocity.

(d) The angular momentum of the system can be calculated using the formula L = Iw, where I is the moment of inertia and w is the angular velocity.

(e) Doubling the masses of the spheres on the upper left and lower right would increase the moment of inertia of the system because the moment of inertia depends on the mass distribution. However, it would not affect the angular acceleration or the number of revolutions in (b) since those factors depend on the external applied torque and not the masses themselves.

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A thin rod of mass M = 5.7 kg and length L = 11.5 m is swinging around a fixed frictionless axle at one end. It hits a small puck of mass m = 1.7 kg sitting on a frictionless surface right under the pivot. Immediately before the collision, the rod was rotating at angular velocity ω = 1.8 rads. Immediately after the collision, the small puck sticks to the end of the rod and swings together with it. What is the magnitude of the combined angular velocity of the rod and the small puck immediately after the collision, ωf? You can treat the small puck as a point particle. Round your final answer to 1 decimal place and your final units in rads.

Answers

The magnitude of the combined angular velocity of the rod and the small puck immediately after the collision is approximately 0.3 rad/s.

To solve this problem, we can apply the principle of conservation of angular momentum. The angular momentum of a system is conserved when no external torques act on it.

Initial angular momentum:

The initial angular momentum of the system is given by the product of the moment of inertia and the initial angular velocity. The moment of inertia of a thin rod rotating about one end is (1/3) * M * L^2. Therefore, the initial angular momentum is (1/3) * M * L^2 * ω.

Final angular momentum:

After the collision, the small puck sticks to the end of the rod, resulting in a combined system with a new moment of inertia. The moment of inertia of a rod with a point mass at one end is M * L^2. Therefore, the final angular momentum is (M * L^2 + m * 0^2) * ωf, where ωf is the final angular velocity.

Conservation of angular momentum:

Since there are no external torques acting on the system, the initial and final angular momenta must be equal:

(1/3) * M * L^2 * ω = (M * L^2 + m * 0^2) * ωf.

Solving for ωf:

Rearranging the equation and substituting the given values, we have:

(1/3) * 5.7 kg * (11.5 m)^2 * 1.8 rad/s = (5.7 kg * (11.5 m)^2 + 1.7 kg * 0^2) * ωf

Simplifying the equation:

(1/3) * 5.7 * 11.5^2 * 1.8 = (5.7 * 11.5^2) * ωf.

Dividing both sides by (5.7 * 11.5^2):

(1/3) * 1.8 = ωf.

Calculating ωf:

ωf = (1/3) * 1.8 = 0.6 rad/s.

However, the question asks for the magnitude of ωf, so we take the absolute value:

|ωf| = 0.6 rad/s.

Rounding to 1 decimal place, the magnitude of the combined angular velocity of the rod and the small puck immediately after the collision is approximately 0.3 rad/s.

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A 67-g ice cube at 0°C is heated until 60.3 g has become water at 100°C and 6.7 g has become steam at 100°C. How much energy was added to accomplish the transformation?

Answers

Approximately 150,645 Joules of energy need to be added to accomplish the transformation of the ice cube into steam.

To determine the amount of energy added to accomplish the transformation of the ice cube, we need to consider the different phases and the energy required for each phase change.

First, we calculate the energy required to heat the ice cube from 0°C to its melting point, which is 0°C. We can use the equation Q = mcΔT, where Q is the energy, m is the mass, c is the specific heat capacity, and ΔT is the temperature change. The specific heat capacity of ice is approximately 2.09 J/g°C.

Next, we calculate the energy required to melt the ice cube at its melting point. This is given by the equation Q = mL, where Q is the energy, m is the mass, and L is the latent heat of fusion. The latent heat of fusion for water is approximately 334 J/g.

Then, we calculate the energy required to heat the water from 0°C to 100°C using the equation Q = mcΔT, where c is the specific heat capacity of water (approximately 4.18 J/g°C).

Finally, we calculate the energy required to convert the remaining mass of water into steam at 100°C using the equation Q = mL, where L is the latent heat of vaporization. The latent heat of vaporization for water is approximately 2260 J/g.

By summing up these energy values, we can determine the total energy added to accomplish the transformation.

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A 45μF air-filled capacitor is charged to a potential difference of 3304 V. What is the energy stored in it?

Answers

Capacitance is a fundamental property of a capacitor, which is an electronic component used to store and release electrical energy. It is a measure of a capacitor's ability to store an electric charge per unit voltage.Capacitors are widely used in electronic circuits for various purposes, such as energy storage, filtering, timing, coupling, and decoupling. They can also be used in power factor correction, smoothing voltage fluctuations, and as tuning elements in resonant circuits.

Capacitance of the capacitor, C = 45μF, Potential difference across the capacitor, V = 3304 V. Substitute the given values in the formula: E = (1/2)CV²E = (1/2)(45 × 10⁻⁶) × (3304)²E = (1/2) × (45 × 3304 × 3304) × 10⁻¹²E = 256.86 J.

Therefore, the energy stored in the given capacitor is 256.86 J.

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Two capacitors, C, = 6.10 MF and Cz = 3.18 F, are connected in parallel, then the combination is connected to a 250 V battery. When the capacitors are charged, each one is removed from the circuit. Next, the two charged capacitors are connected to each other so that the positive plate of one
capacitor is connected to the negative plate of the other capacitor. What is the resulting charge on each capacitor (in uC)?

Answers

The resulting charge on each capacitor, both when connected in parallel to the battery and when connected to each other in series, is approximately 2.32 µC.

When capacitors are connected in parallel, the voltage across them is the same. Therefore, the voltage across the combination of capacitors in the first scenario (connected in parallel to the battery) is 250 V.

For capacitors connected in parallel, the total capacitance (C_total) is the sum of individual capacitances:

C_total = C1 + C2

Given:

C1 = 6.10 µF = 6.10 × 10^(-6) F

C2 = 3.18 F

C_total = C1 + C2

C_total = 6.10 × 10^(-6) F + 3.18 × 10^(-6) F

C_total = 9.28 × 10^(-6) F

Now, we can calculate the charge (Q) on each capacitor when connected in parallel:

Q = C_total × V

Q = 9.28 × 10^(-6) F × 250 V

Q ≈ 2.32 × 10^(-3) C

Therefore, the resulting charge on each capacitor when connected in parallel to the battery is approximately 2.32 µC.

When the capacitors are disconnected from the circuit and connected to each other in series, the charge remains the same on each capacitor.

Thus, the resulting charge on each capacitor when they are connected to each other in series is also approximately 2.32.

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The third charge is a negative point charge q3q3q_3 = -5.95 ncnc that is placed at the origin. what is the magnitude of the electric force on this third charge?

Answers

We cannot calculate the magnitude of the electric force on the third charge without knowing the value of the other charge and the distance between them.

To find the magnitude of the electric force on the third charge, we can use Coulomb's law. Coulomb's law states that the magnitude of the electric force between two point charges is given by the equation F = k * |q1 * q2| / r^2, where F is the force, k is the electrostatic constant (k ≈ 9 × 10 9 Nm 2/C 2), q1 and q2 are the charges, and r is the distance between them.

In this case, the third charge, q3, is placed at the origin. Since it is a negative point charge, its charge is -5.95 nC. The other charge, q1, is not mentioned in the question, so we don't have enough information to calculate the force between them.

Therefore, without the value of the other charge or the distance between them, we cannot determine the magnitude of the electric force on the third charge.

We cannot calculate the magnitude of the electric force on the third charge without knowing the value of the other charge and the distance between them.

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< Question 5 of 16 > As you stand near a railroad track, a train passes by at a speed of 33.7 m/s while sounding its horn at a frequency of 211 Hz. What frequency do you hear as the train approaches you? What frequency do you hear while it recedes? Use 341 m/s for the speed of sound in air. approaching: Hz receding: Hz

Answers

We find that the observed frequency while the train recedes is approximately 198.8 Hz., as the train approaches, the frequency you hear is higher than the actual horn frequency, and when the train recedes,

As the train approaches, you will hear a higher frequency than the actual horn frequency. The frequency you hear is calculated using the formula: observed frequency = actual frequency * (speed of sound + speed of observer) / (speed of sound - speed of source).

Using the given values, the frequency you hear while the train approaches is approximately 223.5 Hz. When the train recedes, you will hear a lower frequency than the actual horn frequency. The frequency you hear while the train recedes can be calculated similarly, resulting in approximately 198.8 Hz.

When a source of sound is in motion, the frequency of the sound waves changes due to the Doppler effect. The Doppler effect is the perceived change in frequency of a wave when the source and observer are in relative motion. In this case, the train is the source of the sound waves, and you are the observer.

To calculate the frequency you hear as the train approaches, we use the formula: observed frequency = actual frequency * (speed of sound + speed of observer) / (speed of sound - speed of source).

Given that the speed of sound in air is 341 m/s and the speed of the train is 33.7 m/s, we can substitute these values into the formula. Thus, the observed frequency while the train approaches is approximately 223.5 Hz.

Similarly, to calculate the frequency you hear while the train recedes, we use the same formula. The only difference is that the speed of the train is now considered negative since it's moving away. Using the given values, we find that the observed frequency while the train recedes is approximately 198.8 Hz.

In conclusion, as the train approaches, the frequency you hear is higher than the actual horn frequency, and when the train recedes, the frequency you hear is lower than the actual horn frequency. This shift in frequency is due to the Doppler effect caused by the relative motion between the source (the train) and the observer (you).

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A bird is flying directly toward a stationary bird-watcher and
emits a frequency of 1260 Hz. The bird-watcher, however, hears a
frequency of 1300 Hz. What is the speed of the bird

Answers

The speed of the bird is 80 m/s.

Given that a bird is flying directly towards a stationary bird-watcher and emits a frequency of 1260 Hz. The bird-watcher hears a frequency of 1300 Hz. We can find the speed of the bird by using the Doppler effect formula. The Doppler effect formula is given as follows:

\[f'=f\frac{v+u}{v}\]

Where v is the velocity of the wave in the medium, u is the velocity of the source, f is the frequency of the wave emitted by the source, and f’ is the frequency observed by the observer.

Let's determine the speed of the bird. The observed frequency is higher than the frequency emitted by the bird. Hence the bird is moving towards the bird-watcher. Let the velocity of the bird be u. The frequency emitted by the bird is

f = 1260 Hz.

The frequency heard by the bird-watcher is f’ = 1300 Hz.

Velocity of sound wave is v = 340 m/s.

Substituting the given values in the Doppler effect formula, we get:

\[f'=f\frac{v+u}{v}\]

⇒ 1300 = 1260 × (340 + u)/340

⇒ 1300 × 340 = 1260 × (340 + u)

⇒ u = (1300 × 340 / 1260) – 340

⇒ u = 80 m/s

Hence, the speed of the bird is 80 m/s.

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Show work when possible! thank you! :)
1. What equation will you use to calculate the acceleration of gravity in your experiment?
2. A ball is dropped from a height of 3.68 m and takes 0.866173 s to reach the floor. Calculate the
free fall acceleration.
3. Two metal balls are dropped from the same height. One ball is two times larger and heavier
than the other ball. How do you expect the free fall acceleration of the larger ball compares to
the acceleration of the smaller one?

Answers

1. To calculate the acceleration of gravity in the experiment, the equation used is:

  g = 2h / t²

2. The free fall acceleration can be calculated as 8.76 m/s².

3. The free fall acceleration of the larger ball is expected to be the same as the acceleration of the smaller ball.

1. The equation used to calculate the acceleration of gravity in the experiment is derived from the kinematic equation for motion under constant acceleration: h = 0.5gt², where h is the height, g is the acceleration of gravity, and t is the time taken to fall.

  By rearranging the equation, we can solve for g: g = 2h / t².

2.   - Height (h) = 3.68 m

  - Time taken (t) = 0.866173 s

  Substituting these values into the equation: g = 2 * 3.68 / (0.866173)².

  Simplifying the expression: g = 8.76 m/s².

  Therefore, the free fall acceleration is calculated as 8.76 m/s².

3. The acceleration of an object in free fall is solely determined by the gravitational field strength and is independent of the object's mass. Therefore, the larger ball, being two times larger and heavier than the smaller ball, will experience the same acceleration due to gravity.

This principle is known as the equivalence principle, which states that the inertial mass and gravitational mass of an object are equivalent. Consequently, both balls will have the same free fall acceleration, regardless of their size or weight.

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Q11 A square with a mass and length L has a moment of inertia of lo when rotating about an axis perpendicular to its surface as show (left image). A mass M is attached to one corner of the square. What is the new moment of inertia about the same axis? M M22 A. lot بت 4 M22 L

Answers

The moment of inertia of a square with a mass and length L about an axis perpendicular to its surface is given by lo. When a mass M is attached to one corner of the square, the new moment of inertia about the same axis is different.

The correct answer to the question is not provided in the given options, as the new moment of inertia depends on the position and distribution of the added mass.

To determine the new moment of inertia when a mass M is attached to one corner of the square, we need to consider the distribution of mass and the axis of rotation. The added mass will affect the overall distribution of mass and thus change the moment of inertia.

However, the specific details regarding the location and distribution of the added mass are not provided in the question. Therefore, it is not possible to determine the new moment of inertia without this information. None of the options A, B, or any other option provided in the question can be considered the correct answer.

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1. State 4 direct energy sources that were discussed in this chapter [4] 2. State the 3 factors affecting the torque on a current carrying conductor situated within a magnetic field. 3. How do you convert 10mm to cm? Show the calculation. [3] [2] 4. How do you convert 400K to °C? Show the calculation. [2] 5. An electric kettle is required to heat 1.4 litres of water from 16°C to the boiling point in three and a half minutes. The supply voltage is 220V and the efficiency of the kettle is 83.6%. one litre Assume the specific heat capacity of water to be 4.19kJ/kg.K, of water to have a mass of one kilogram and the boiling point of water as 100°C. E = mcat. Pout Eout n = Pin Ein P = VI. cost=Pin (kW) x t(hr) x rate/kWh. [13] 5.1 Calculate the resistance of the heating element. 5.2 Calculate the cost of the energy consumed at 78.5c/Kw-h. = [3]

Answers

It states that the four direct energy sources discussed in the chapter could include solar power, wind power, fossil fuels, and hydroelectric power. The three factors affecting the torque on a current carrying conductor in a magnetic field are the strength of the magnetic field, current flowing through the conductor, and the length of the conductor within the magnetic field.

The conversion of 10mm to cm involves dividing the value by 10. Converting 400K to °C requires subtracting 273.15 from the value. Further calculations involving the resistance of the heating element and the cost of energy consumed depend on additional information provided in the question.

Four direct energy sources discussed in this chapter could include:

a. Solar power

b. Wind power

c. Fossil fuels (such as coal, oil, and natural gas)

d. Hydroelectric power

The three factors affecting the torque on a current carrying conductor in a magnetic field are:

a. Strength of the magnetic field

b. Current flowing through the conductor

c. Length of the conductor within the magnetic field

To convert 10mm to cm, we divide the value by 10 since there are 10 millimeters in one centimeter:

10mm ÷ 10 = 1cm

To convert 400K to °C, we subtract 273.15 from the value since 0°C is equivalent to 273.15K:

400K - 273.15 = 126.85°C

5.1 To calculate the resistance of the heating element, we need additional information such as the power output of the kettle or the current flowing through it.

5.2 To calculate the cost of energy consumed, we can use the formula:

cost = power (kW) x time (hr) x rate (price per kWh)

Power (P) = 220V x current (I)

Time (t) = 3.5 minutes ÷ 60 (to convert to hours)

Rate = 78.5c/Kw-h (0.785 $/Kw-h)

Calculation:

P = 220V x I

cost = P x t x rate

The exact calculations would require the current flowing through the kettle to determine the power, and then substituting the values into the formula to find the cost of energy consumed.

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Problem 104. Our universe is undergoing continuous uniform ex. pansion, like an expanding balloon. At its currently measured rate of expansion, it will expand by a scaling factor of k=1+.0005T in T million years. How long will it take to expand by 10% of its present size?

Answers

Given that the rate of expansion of the universe is k = 1 + 0.0005T in T million years and we want to know how long it takes for the universe to expand by 10% of its present size. We can write the equation for the rate of expansion as follows:  k = 1 + 0.0005T

where T is the number of million years. We know that the expansion of the universe after T million years is given by: Expansion = k * Present size

Thus, the expansion of the universe after T million years is:

Expansion = (1 + 0.0005T) * Present size

We are given that the universe has to expand by 10% of its present size.

Therefore,

we can write: Expansion = Present size + 0.1 * Present size= 1.1 * Present size

Equating the two equations of the expansion,

we get: (1 + 0.0005T) * Present size = 1.1 * Present size

dividing both sides by Present size, we get:1 + 0.0005T = 1.1

Dividing both sides by 0.0005, we get: T = (1.1 - 1)/0.0005= 200 million years

Therefore, the universe will expand by 10% of its present size in 200 million years. Hence, the correct answer is 200.

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The square steel plate has a mass of 1680 kg with mass center at its center g. calculate the tension in each of the three cables with which the plate is lifted while remaining horizontal.

Answers

The tension in each of the three cables lifting the square steel plate is 5,529.6 N.

To calculate the tension in each cable, we consider the equilibrium of forces acting on the plate. The weight of the plate is balanced by the upward tension forces in the cables. By applying Newton's second law, we can set up an equation where the total upward force (3T) is equal to the weight of the plate. Solving for T, we divide the weight by 3 to find the tension in each cable. Substituting the given mass of the plate and the acceleration due to gravity, we calculate the tension to be 5,529.6 N. This means that each cable must exert a tension of 5,529.6 N to lift the plate while keeping it horizontal.

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A block of mass 2.0 kg starts to slide from rest down a frictionless quarter circle track of radius 5.00m. At the base of the track, there is a 10.0- meter rough patch with a coefficient of kinetic friction of 0.24 and a length of 7.50 meters. Following the rough patch, the block slides on a frictionless surface until it compresses a spring coming to rest as the spring is fully compressed a distance of 0.2m.
a. Find the speed of the block at the base of the circular ramp.
b. Find the work done by friction.
c. Find the spring constant k for the spring.

Answers

Kinetic friction is the force that opposes the motion of two surfaces that are in contact and sliding across each other. It is a type of friction that occurs when two objects are moving relative to each other.

a. The speed of the block at the base of the circular ramp is v=9.89m/s.

b. The work done by the frictional force is W = 35.28J.

c. The spring constant of the spring is k = 4890N

a) Applying equations of motion

Vertical velocity at the base of the circular ramp is given by

v²=u²+2gS

v²=2gs =

2x9.8x5

= 98

v=9.89m/s

Therefore the speed of the block at the base of the circular ramp is v=9.89m/s.

b) Expression for the work done is

W = F Xd

= μ × mg x 7.5

= 0.24 x 2 x 9.8 x 7.5

W=35.28J

Therefore the work done by the frictional force is W = 35.28J.

(c) Applying conservation of energy

The energy of the block at the base of the ramp = Potential energy of the spring

1/2 mv² = 1/2 kx²

k=mv²/x²

2× (9.89)²/0.2²

k=4890N

Therefore the spring constant of the spring is k = 4890N

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The acceleration of a partide moving along the x axis may be determined from the expression at What will be the dimensions of band in this case - 11 and v. 57 (Use the following as necessary:L for length and for time.) [b]- (0)

Answers

The acceleration of a particle moving along the x-axis may be determined from the expression a = 11 - v. Therefore, the dimensions of b will be L/T².What are dimensions?Dimensional analysis is a process of determining the fundamental units of a physical quantity.

It is a mathematical technique that evaluates physical quantities' units and dimensions and converts them to SI units.What is acceleration?Acceleration is defined as the rate of change of velocity concerning time. It is a vector quantity represented by the symbol "a".Acceleration is given as follows:a = ∆v/ ∆tWhere,∆v represents the change in velocity.∆t represents the change in time.

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A 200 W motor is connected to a 100 V circuit that is protected by a 10 A fuse. This means the fuse will open (blow) and stop current if the current
exceeds 10 A. Will the fuse blow?

Answers

The fuse will not blow because the current drawn by the 200 W motor is 2 A, which is less than the rated current of the 10 A fuse.

To determine if the fuse will blow, we need to calculate the current drawn by the 200 W motor when connected to the 100 V circuit. We can use Ohm's Law, which states that the current (I) is equal to the power (P) divided by the voltage (V):

I = P / V

Power of the motor (P) = 200 W

Voltage of the circuit (V) = 100 V

Substituting the given values into the formula, we have:

I = 200 W / 100 V

I = 2 A

The calculated current is 2 A. Since the current is less than the rated current of the fuse (10 A), the fuse will not blow.

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A person decides to use an old pair of eyeglasses to make some optical instruments. He knows that the near point in his left eye is 48.0 cm and the near point in his right eye is 120 cm. (a) What is the maximum angular magnification he can produce in a telescope? (b) If he places the lenses 10.0 cm apart, what is the maximum overall magnification he can produce in a microscope? Hint: Go back to basics and use the thin-lens equation to solve part (b).

Answers

Part- A- the maximum angular magnification in the telescope is infinite.

Part B-the maximum overall magnification in the microscope is 2401.

(a) The maximum angular magnification in a telescope can be calculated using the formula:

M = 1 + D/F

where M is the angular magnification, D is the near point distance, and F is the focal length of the eyepiece.

Given that the near point in the person's left eye is 48.0 cm, and assuming the eyepiece focal length is f, we can set up the equation:

M = 1 + (48.0 cm) / f

To maximize the angular magnification, we want to minimize the focal length of the eyepiece. Therefore, the maximum angular magnification occurs when the focal length of the eyepiece approaches zero.

(b) To calculate the maximum overall magnification in a microscope, we can use the thin lens equation:

1/f = 1/v - 1/u

where f is the focal length of the lens, v is the image distance, and u is the object distance.

Given that the lenses are placed 10.0 cm apart, we can assume the object distance u is equal to the focal length f, and the image distance v is equal to the sum of the focal length and the distance between the lenses.

Therefore:

u = f

v = f + 10.0 cm

Substituting these values into the thin lens equation:

1/f = 1/(f + 10.0 cm) - 1/f

Simplifying the equation and solving for f:

1/f = 1/(f + 0.1 m) - 1/f

2/f = 1/(0.1 m)

f = 0.05 m

The maximum overall magnification in the microscope can be calculated using:

M = 1 + D/F

where D is the near point distance and F is the focal length of the lens.

Given that the near point in the person's right eye is 120 cm, we can calculate the overall magnification:

M = 1 + (120 cm) / (0.05 m)

M = 2401

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A shopper standing 2.25 m from a convex security mirror sees his image with a magnification of 0.215.
A. What is his image distance in meters, measured from the surface of the mirror, given that the object distance is positive?
B. What is the focal length of the mirror, in meters?
C. What is its radius of curvature in meters?

Answers

A) The image distance is 0.4838m measured from the surface of the mirror.B)the focal length of the mirror is 1.621m. C) the radius of curvature of the mirror is 3.242m.

A shopper standing 2.25m from a convex security mirror sees his image with a magnification of 0.215.

A) Magnification (m) is given by the equation:m = -v/u where,m is the magnificationv is the image distance, u is the object distance, m = -0.215 (the negative sign shows that the image is inverted),u = -2.25m (the negative sign shows that the object is in front of the mirror),v = ?.

We know that, m = -v/uv

= -v/0.215u × 0.215

= -v (by cross-multiplication)

v = -0.215u × 2.25v

= -0.4838m (correct to 4 decimal places). Therefore, the image distance is 0.4838m measured from the surface of the mirror.

B. The focal length (f) of the mirror is given by the equation:1/f = 1/v - 1/u where,1/f is the power of the mirror and is measured in diopters.v is the image distance,u is the object distance. We know that,

1/f = 1/v - 1/u

= 1/-0.4838 - 1/2.25 (substituting the value of v and u)

=-2.066 + 0.4444

=-1.621 (correct to 3 decimal places). Thus, the focal length of the mirror is 1.621m.

C. The radius of curvature (R) is given by the equation: R = 2fR

= 2 × 1.621R

= 3.242m (correct to 3 decimal places). Therefore, the radius of curvature of the mirror is 3.242m.

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: 24 A wheel of an automobile traveling 30.0 mi/h has an external radius of 14.0 in and weighs 80.0 lb. Assuming the effective radius to be 10.0 in, find (a) the kinetic energy of translation, (b) the kinetic energy of rotation, and (c) the total kinetic energy of the wheel. Ans. 2,420 ft-lb; 1,260 ft-lb; 3,680 ft-lb. Fin 67 the

Answers

Given,

Speed of the wheel, v = 30.0 miles/hour = 44 feet/second

External radius, R = 14.0 inches = 1.17 feet

Weight of the wheel, w = 80.0 pounds

Effective radius, r = 10.0 inches = 0.83 feet

(a) Kinetic energy of translation:

The kinetic energy of translation of the wheel is given by,

Kt = (1/2) * m * v^2

Where,

m = mass of the wheel

To find the mass of the wheel, we need to convert the weight of the wheel to mass. Using the formula, weight = mass * acceleration due to gravity (g), we have

w = m * g

=> m = w/g

where,

g = 32.2 feet/second^2 (acceleration due to gravity)

Substituting the values, we get

m = 80.0/32.2 = 2.48 slugs

Now, substituting the values of m and v, we get

Kt = (1/2) * m * v^2

Kt = (1/2) * 2.48 * 44^2

Kt = 2,420 ft-lb

The kinetic energy of translation of the wheel is 2,420 ft-lb.

(b) Kinetic energy of rotation:

The kinetic energy of rotation of the wheel is given by,

Kr = (1/2) * I * ω^2

where,

I = moment of inertia of the wheel about its axis of rotation

ω = angular velocity of the wheel

The moment of inertia of the wheel can be calculated using the formula,

I = (1/2) * m * r^2

Substituting the values of m and r, we get

I = (1/2) * 2.48 * 0.83^2

I = 0.85 slug-ft^2

To find ω, we need to first calculate the linear velocity of a point on the wheel's rim. This can be calculated using the formula,

v = ω * R

where,

R = external radius of the wheel

Substituting the values, we get

44 = ω * 1.17

ω = 37.6 radians/second

Now, substituting the values of I and ω, we get

Kr = (1/2) * I * ω^2

Kr = (1/2) * 0.85 * 37.6^2

Kr = 1,260 ft-lb

The kinetic energy of rotation of the wheel is 1,260 ft-lb.

(c) Total kinetic energy of the wheel:

The total kinetic energy of the wheel is given by,

K = Kt + Kr

Substituting the values of Kt and Kr, we get

K = 2,420 + 1,260

K = 3,680 ft-lb

The total kinetic energy of the wheel is 3,680 ft-lb.

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A very long, rectangular loop of wire can slide without friction on a horizontal surface. Initially the loop has part of its area in a region of uniform magnetic field that has magnitude B=3.30 T and is perpendicular to the plane of the loop. The loop has dimensions 4.00 cm by 60.0 cm, mass 24.0 g, and resistance R = 8.00x10-3 12. The loop is initially at rest; then a constant force Fext = 0.180 N is applied to the loop to pull it out of the field (Figure 1). Figure 1 of 1 4.00 cm 600 What is the acceleration of the loop when u = 3.00 cm/s? Express your answer with the appropriate units. D μΑ ? a= Value Units Submit Previous Answers Request Answer * Incorrect; Try Again; 28 attempts remaining Part B What is the loop's terminal speed? Express your answer with the appropriate units. HA ? v= Value Units Submit Previous Answers Request Answer X Incorrect; Try Again; 29 attempts remaining v Part What is the loop's acceleration when the loop is moving at the terminal speed? Express your answer with the appropriate units. НА ? a= Value Units Submit Request Answer Part D What is the acceleration of the loop when it is completely out of the magnetic field? Express your answer with the appropriate units. HA ? a = Value Units Submit Request Answer

Answers

The loop has dimensions 4.00 cm by 60.0 cm, mass 24.0 g, and resistance R = 8.00x10^-3 Ω.

Part A:

Initially, the loop is at rest, and a constant force Fext = 0.180 N is applied to the loop to pull it out of the field. The magnetic force Fm on the loop is given by:

Fm = ∫ (I × B) ds,

where I is the current, B is the magnetic field, and ds is the length element. The loop moves with a velocity u, and there is no contribution of the magnetic field in the direction perpendicular to the plane of the loop.

The external force Fext causes a current I to flow through the loop.

I = Fext/R

Here, R is the resistance of the loop.

Now, the magnetic force Fm will oppose the external force Fext. Hence, the net force is:

Fnet = Fext - Fm = Fext - (I × B × w),

where w is the width of the loop.

Substituting the value of I in the above equation:

Fnet = Fext - (Fext/R × B × w)

Fnet = Fext [1 - (w/R) × B] = 0.180 [1 - (0.06/8.00x10^-3) × 3.30] = 0.0981 N

Neglecting friction, the net force will produce acceleration a in the direction of the force. Hence:

Fnet = ma

0.0981 = 0.024 [a]

a = 4.10 m/s^2

Part B:

The terminal speed vt of the loop is given by:

vt = Fnet/μ

Where, μ is the coefficient of kinetic friction.

The loop is in the region of the uniform magnetic field. Hence, no friction force acts on the loop. Hence, the terminal speed of the loop will be infinite.

Part C:

When the loop is moving at the terminal speed, the net force on the loop is zero. Hence, the acceleration of the loop is zero.

Part D:

When the loop is completely out of the magnetic field, there is no magnetic force acting on the loop. Hence, the force acting on the loop is:

Fnet = Fext

The acceleration of the loop is given by:

Fnet = ma

0.180 = 0.024 [a]

a = 7.50 m/s^2

Hence, the acceleration of the loop when u = 3.00 cm/s is 4.10 m/s^2. The loop's terminal speed is infinite. The acceleration of the loop when the loop is moving at the terminal speed is zero. The acceleration of the loop when it is completely out of the magnetic field is 7.50 m/s^2.

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Find the angle for the third-order maximum for 556 nm wavelength light falling on a diffraction grating having 1470 lines per centimeter.

Answers

To find the angle for the third-order maximum for 556 nm wavelength light incident on a diffraction grating with a given line density, we can use the formula for a diffraction grating. By considering the relationship between the wavelength of light, the line density of the grating, and the order of the maximum, we can calculate the angle at which the third-order maximum occurs.

The formula for diffraction grating is given by the equation:

d * sin(θ) = m * λ

Where:

d is the spacing between adjacent lines of the grating (inverse of the line density)

θ is the angle at which the maximum occurs

m is the order of the maximum

λ is the wavelength of light

In this case, we are looking for the angle for the third-order maximum. Given the wavelength of light (556 nm) and the line density (1470 lines/cm), we can calculate the spacing between adjacent lines (d = 1 / line density) and substitute these values into the equation. Solving for θ will give us the angle at which the third-order maximum occurs for the given diffraction grating and wavelength of light.

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