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2 years ago

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A conducting rod (length = 80 cm) rotates at a constant angular rate of 15 revolutions per second about a pivot at one end. A un

iform field (B = 60 mT) is directed perpendicularly to the plane of rotation. What is the magnitude of the emf induced between the ends of the rod?

1 answer:

Studentka2010 [4]2 years ago

4 0

Answer:

3.62 V

Explanation:

L = 80 cm = 0.8 m

f = 15 rps

B = 60 m T = 0.060 T

ω = 2 x π x f = 2 x 3.14 x 15 = 94.2 rad/s

v = r ω

here, r be the radius of circular path. Here r = length of rod = L

v = 0.80 x 94.2 = 75.36 m/s

The motional emf is given by

e = B v L = 0.060 x 75.36 x 0.8 = 3.62 V

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The number of gallons of water in a storage tank at time t, in minutes, is modeled by w(t) = 25 − t2 for 0≤t≤5. At what rate, in

Kisachek [45]

Answer:

Rate of change of water will be -6 gallon/minute

Explanation:

We have given water in the tank as the function of time as

$w(t)=25-t^2$

We have to find the rate of change of water in the tank at t = 3 minute

For rate of change we have to differentiate both side

So $\frac{dw}{dt}=0-2t$

At t = 3 minute

$\frac{dw}{dt}=0-2\times 3=-6gallon/minute$

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2 years ago

un tanque de gasolina de 40 litros fue llenado por la noche, cuando la temperatura era de 68 grados farenheit. Al dia siguiente

Sedaia [141]

Answer:

Volume of gasoline that expands and spills out is 1.33 ltr

Explanation:

As we know that when temperature of the liquid is increased then its volume will expand and it is given as

$\Delta V = V_o\gamma \Delta T$

here we know that

$V_o = 40 Ltr$

volume expansion coefficient of the gasoline is given as

$\gamma = 950 × 10^{–6}$

change in temperature is given as

$\Delta T = (131 - 68) \times \frac{5}{9}$

$\Delta T = 35 ^oC$

Now we have

$\Delta V = 40(950 \times 10^{-6})(35)$

$\Delta V = 1.33 Ltr$

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2 years ago

A high school physics instructor catches one of his students chewing gum in class. He decides to discipline the student by askin

KengaRu [80]

a) 219.8 rad/s

b) $20.0 rad/s^2$

c) $2.9 m/s^2$

d) $7005 m/s^2$

e) Towards the axis of rotation

f) $0 m/s^2$

g) 31.9 m/s

Explanation:

a)

The angular velocity of an object in rotation is the rate of change of its angular position, so

$\omega=\frac{\theta}{t}$

where

$\theta$ is the angular displacement

t is the time elapsed

In this problem, we are told that the maximum angular velocity is

$\omega_{max}=35 rev/s$

The angle covered during 1 revolution is

$\theta=2\pi$ rad

Therefore, the maximum angular velocity is:

$\omega_{max}=35 \cdot 2\pi = 219.8 rad/s$

b)

The angular acceleration of an object in rotation is the rate of change of the angular velocity:

$\alpha = \frac{\Delta \omega}{t}$

where

$\Delta \omega$ is the change in angular velocity

t is the time elapsed

Here we have:

$\omega_0 = 0$ is the initial angular velocity

$\omega_{max}=219.8 rad/s$ is the final angular velocity

t = 11 s is the time elapsed

Therefore, the angular acceleration is:

$\alpha = \frac{219.8-0}{11}=20.0 rad/s^2$

c)

For an object in rotation, the acceleration has two components:

- A radial acceleration, called centripetal acceleration, towards the centre of the circle

- A tangential acceleration, tangential to the circle

The tangential acceleration is given by

$a_t = \alpha r$

where

$\alpha$ is the angular acceleration

r is the radius of the circle

Here we have

$\alpha =20.0 rad/s^2$

d = 29 cm is the diameter, so the radius is

r = d/2 = 14.5 cm = 0.145 m

So the tangential acceleration is

$a_t=(20.0)(0.145)=2.9 m/s^2$

d)

The magnitude of the radial (centripetal) acceleration is given by

$a_c = \omega^2 r$

where

$\omega$ is the angular velocity

r is the radius of the circle

Here we have:

$\omega_{max}=219.8 rad/s$ is the angular velocity when the fan is at full speed

r = 0.145 m is the distance of the gum from the centre of the circle

Therefore, the radial acceleration is

$a_c=(219.8)^2(0.145)=7005 m/s^2$

e)

The direction of the centripetal acceleration in a rotational motion is always towards the centre of the axis of rotation.

Therefore also in this case, the direction of the centripetal acceleration is towards the axis of rotation of the fan.

f)

The magnitude of the tangential acceleration of the fan at any moment is given by

The tangential acceleration is given by

$a_t = \alpha r$

where

$\alpha$ is the angular acceleration

r is the radius of the circle

When the fan is rotating at full speed, we have:

$\alpha=0$ , since the fan is no longer accelerating, because the angular velocity is no longer changing

r = 0.145 m

Therefore, the tangential acceleration when the fan is at full speed is

$a_t=(0)(0.145)=0 m/s^2$

g)

The linear speed of an object in rotational motion is related to the angular velocity by the formula:

$v=\omega r$

where

v is the linear speed

$\omega$ is the angular velocity

r is the radius

When the fan is rotating at maximum angular velocity, we have:

$\omega=219.8 rad/s$

r = 0.145 m

Therefore, the linear speed of the gum as it is un-stucked from the fan will be:

$v=(219.8)(0.145)=31.9 m/s$

7 0

2 years ago

A spring is used to launch a coffee mug. The 20cm long spring can be compressed by a maximum of 8cm. The mug has a mass of 350 g

Zepler [3.9K]

Answer:

7503.13 N/m

Explanation:

Use principle of conservation of energy.

Here, energy stored in the spring due to compression shall be utilized in attaining the potential energy of the mug.

Given that,

Length of the spring = 20 cm = 0.20 m

Compression, x = 8 cm = 0.08 m

mass of the mug, m = 350 g = 0.35 kg

h = 7 m

use the expression for energy balance -

(1/2)*k*x^2 = m*g*h

=> k = (2*m*g*h) / x^2

input the values

k = (2*0.35*9.8*7) / 0.08^2

= 7503.13 N/m

8 0

2 years ago

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A hummingbird 3.4m above the ground flies 1.2 m along a straight line path. Upon spotting a flower below, the hummingbird drops

Anit [1.1K]

A = horizontal displacement of the humming bird = 1.2 m

B = vertical displacement of the humming bird = 1.4 m

C = net displacement of the humming bird from initial to final position = ?

In the triangle drawn , Using Pythagorean theorem

C = √(A² + B²)

inserting the values

C = √(1.2² + 1.4²)

C = √(1.44 + 1.96)

C = √(3.4)

C = 1.4 m

Hence the net displacement of hummingbird comes out to be 1.4 m

4 0

2 years ago

Read 2 more answers