All categories

3 years ago

Which description correctly summarizes how an electric motor causes an axle to turn?

Physics

1 answer:

alexira [117]3 years ago

6 0

Answer:

B on edge

Explanation:

the piles of the magnetic field generated around the armature are attracted to the opposite poles of the permanent magnent. As the opposite poles align, the commutator reverses the current direction so like poles are aligned and the armature continues to spin

You might be interested in

. A block is placed on an incline. The coefficient of static friction between the block and the plane is 0.59. What is the maxim

pychu [463]

Answer:

$\theta = 30.5 degree$

Explanation:

If the block remains in equilibrium on the inclined plane

then we can say that component of weight along the inclined plane is counter balanced by the static friction force.

Here we know that maximum static friction force on the block is given by

$f_s = \mu_s F_n$

here we have

$F_n = mgcos\theta$

now the component of weight along the inclined is balance by maximum static friction force

$\mu_s (mgcos\theta) = mg sin\theta$

$tan\theta = \mu_s$

$tan\theta = 0.59$

$\theta = tan^{-1}(0.59)$

$\theta = 30.5 degree$

3 0

3 years ago

What is the definition of half metals or semi metals?

Lyrx [107]

Those metals which conduct the electricity or heat partially, which may be good or bad conductors of heat and electricity is half -metals or semi metals.

7 0

3 years ago

Read 2 more answers

- Thermal Energy Test

timofeeve [1]

Answer:D. The metal in the door expands when it becomes warm due to an increase in kinetic energy, and contracts when it's cooler.

Explanation:

because the metal in the door expands due to warm to an increase in kinetic energy and contracts.

4 0

3 years ago

A 0.42 kg shuffleboard disk is initially at rest when a player uses a cue to increase its speed to 4.2 m/s at constant accelerat

MariettaO [177]

Answer

given,

mass of the shuffleboard disk = 0.42 kg

speed of the cue is increased to = 4.2 m/s

acceleration takes over 2 m then acceleration is zero.

the disk additionally slide to 12 m

final speed of disk = 0 m/s

a) increase in thermal energy

$\Delta E_t = \dfrac{1}{2}m(v_1^2-v_2^2)$

$\Delta E_t = \dfrac{1}{2}\times 0.42 \times (4.2^2-v_2^2)$

$\Delta E_t = 3.704\ J$

b) $\Delta E_t = F_f.d$

F_f is the frictional force

$3.704= 12.F$

$F_f = 0.308\ N$

increase in thermal energy for entire movement of 14 m

$\Delta E_t = 0.308\times 14$

$\Delta E_t =4.312 \ J$

c) Work done on the disk by the cue

$W = \Delta KE + \Delta E_{t}$

$W = \dfrac{1}{2}\times 0.42 \times (4.2^2-0^2)+ F_f \times d$

$W = 3.704+ 0.308 \times 2$

W = 3.704 + 0.616

W = 4.32 J

3 0

4 years ago

An electric motor can drive grinding wheel at two different speeds. When set to high the angular speed is 2000 rpm. The wheel tu

shutvik [7]

a) The initial angular speed is 209.3 m/s

b) The angular acceleration is $-1.74 rad/s^2$

c) The angular speed after 40 s is 139.7 rad/s

d) The wheel makes 1501 revolutions

Explanation:

The initial angular speed of the wheel is

$\omega_i = 2000 rpm$

which means 2000 revolutions per minute.

We have to convert it into rad/s. Keeping in mind that:

$1 rev = 2\pi rad$

$1 min = 60 s$

We find:

$\omega_i = 2000 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=209.3 rad/s$

To find the angular acceleration, we have to convert the final angular speed also from rev/min to rad/s.

Using the same procedure used in part a),

$\omega_f = 1000 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=104.7 rad/s$

Now we can find the angular acceleration, given by

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_i = 209.3 rad/s$ is the initial angular speed

$\omega_f = 104.7 rad/s$ is the final angular speed

t = 60 s is the time interval

Substituting,

$\alpha = \frac{104.7-209.3}{60}=-1.74 rad/s^2$

To find the angular speed 40 seconds after the initial moment, we use the equivalent of the suvat equations for circular motion:

$\omega' = \omega_i + \alpha t$

where we have

$\omega_i = 209.3 rad/s$

$\alpha = -1.74 rad/s^2$

And substituting t = 40 s, we find

$\omega' = 209.3 + (-1.74)(40)=139.7 rad/s$

The angular displacement of the wheel in a certain time interval t is given by

$\theta=\omega_i t + \frac{1}{2}\alpha t^2$

where

$\omega_i = 209.3 rad/s$

$\alpha = -1.74 rad/s^2$

And substituting t = 60 s, we find:

$\theta=(209.3)(60) + \frac{1}{2}(-1.74)(60)^2=9426 rad$

So, the wheel turns 9426 radians in the 60 seconds of slowing down. Converting this value into revolutions,

$\theta = \frac{9426 rad}{2\pi rad/rev}=1501 rev$

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

8 0

3 years ago