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Scorpion4ik [409]
1 year ago
6

Which changes in an electric motor will make the motor stronger? Check all that apply. using a stronger permanent magnet using a

weaker permanent magnet increasing current in the electromagnet decreasing current in the electromagnet increasing the distance between the magnets decreasing the distance between the magnets
Physics
2 answers:
wariber [46]1 year ago
7 0
<h2>Answer:</h2>

The following changes will make the motor more stronger

  • <u>Using strong permanent magnets</u>
  • <u>Increasing the current</u>
  • <u>Decreasing the space between magnets</u>
<h2>Explanation:</h2>

Using string magnets will produce strong flux and the motor will become strong. The internal flux also deepness upon the strength of coil. Similarly by increasing current we can make the motor more powerful because according to power equation P=VI so increasing the current will increase the product of voltage and current resulting more power. In the same way by reducing the distance between magnets will make the motor stronger because the flux and torque will increase.

miv72 [106K]1 year ago
4 0
Use stronger magnets
increase current
push magnets closer to coil
adding more sets of coils
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A high school physics instructor catches one of his students chewing gum in class. He decides to discipline the student by askin
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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

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Given that,

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Answer:

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