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gulaghasi [49]
4 years ago
15

In an electric motor, a commutator

Physics
2 answers:
alexdok [17]4 years ago
4 0

Answer:

b. Repeatedly reverses the flow of current through the armature.

Explanation:

  1. The electric motor is a device that converts the electrical energy into mechanical energy.
  2. The commutator is a device that repeatedly reverses the flow of current through the armature.
  3. It is a split ring that acts as the commutator.
  4. It connects the external circuit through a carbon brush and the armature of the motor.
  5. Another important component of an electric motor is an electromagnet or permanent magnet.
Monica [59]4 years ago
4 0

Answer:

b on edg 2020

Explanation:

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In a "worst-case" design scenario, a 2000-{\rm kg} elevator with broken cables is falling at 4.00{\rm m/s} when it first contact
OLga [1]

Answer:

v=3.73m/s,   a=3.35m/s^{2}

Explanation:

In order to solve this problem, we must first draw a diagram that will represent the situation (See attached picture).

So there are two ways in which we can solve this proble. One by using integrls and the other by using energy analysis. I'll do it by analyzing the energy of the system.

So first, we ned to know what the constant of the spring is, so we can find it by analyzing the energy on points A and C. So we get the following:

K_{A}+U_{Ag}=K_{C}+U_{Cs}+U_{Cg}+W_{Cf}

where K represents kinetic energy, U represents potential energy and W represents work.

We know that the kinetic energy in C will be zero because its speed is supposed to be zero. We also know the potential energy due to the gravity in C is also zero because we are at a height of 0m. So we can simplify the equation to get:

K_{A}+U_{Ag}=U_{Cs}+W_{Cf}

so now we can use the respective formulas for kinetic energy and potential energy, so we get:

\frac{1}{2}mV_{A}^{2}+mgh_{A}=\frac{1}{2}ky_{C}+fy_{C}

so we can now solve this for k, which is the constant of the spring.

k=\frac{mV_{A}^{2}+mgh_{A}-fy_C}{y_{C}^{2}}

and now we can substitute the respective data:

k=\frac{(2000kg)(4m/s)^{2}+(2000kg)(9.8m/s^{2})(2m)-(17000N)(2m)}{(2m)^{2}}

Which yields:

k= 9300N/m

Once we got the constant of the spring, we can now find the speed of the elevator at point B, which is one meter below the contact point, so we do the same analysis of energies, like this:

K_{A}+U_{Ag}=K_{B}+U_{Bs}+U_{Bg}+W_{Bf}

In this case ther are no zero values to eliminate so we work with all the types of energies involved in the equation, so we get:

\frac{1}{2}mV_{A}^{2}+mgh_{A}=\frac{1}{2}mV_{B}^{2}+\frac{1}{2}ky_{B}^{2}+mgh_{B}+fy_{B}

and we can solve for the Velocity in B, so we get:

V_{B}=\sqrt{\frac{mV_{A}^{2}-ky_{B}^{2}+2mgh_{B}-2fy_{B}}{m}}

we can now input all the data directly, so we get:

V_{B}=\sqrt{\frac{(2000kg)(4m/s)^{2}-(9300N/m)(1m)^{2}+2(2000kg)(9.8m/s^{2})(1m)-2(17000N)(1m)}{(2000kg)}}

which yields:

V_{B}=3.73m/s

which is the first answer.

Now, for the second answer we need to find the acceleration of the elevator when it reaches point B, for which we need to build a free body diagram (also included in the attached picture) and to a summation of forces:

\sum F=ma

so we get:

F_{s}+f-W=ma

Supposing the acceleration is positive when it points upwards.

So we can solve that for the acceleration, so we get:

a=\frac{F_{s}+f-W}{m}

or

a=\frac{ky_{B}+f-W}{m}

and now we substitute the data we know:

a=\frac{(9300N/m)(1m)+(17000N)-(2000kg)(9.8m/s^{2})}{2000kg}

which yields:

3.35m/s^{2}

8 0
3 years ago
Which is a characteristic of the image formed between F and the center of the lens?
m_a_m_a [10]

Answer: The image is upright

Explanation:

The lens shown in the diagram is concave lens. It is a diverging lens which means that the light rays which fall from the object onto the lens diverge.

Here, the object is kept between focus and center of lens. The image would form between focus and center of the lens at the same side of the object. The image would virtual, diminished and upright. Thus, the correct option is: The image is upright.

3 0
3 years ago
When a machine is used to do work, the force applied by the machine is called the effort force?
nignag [31]
Y e s  i t ' s   c a l l e d  "  t h e   e f f o r t   f o r c e  " 

H o p e  t h i s  h e l p s

:) L O L 
8 0
3 years ago
A stone is thrown vertically upward with a speed of 12m/s from the edge of a cliff 70 m high (a) How much later it reaches the b
jonny [76]

Answer

given,

vertical speed of stone,v = 12 m/s

height of the cliff = 70 m

a) time taken by the stone to reach at the bottom of the cliff

We know that,

S = u t + 1/2 a t²

- 70 = 12 t - 0.5 x 9.8 t²

4.9 t² - 12 t - 70 = 0  

solving the equation

t = 5.2 s (neglecting the negative value)

b) again using equation of motion

   v = u + a t

   v = 12 - 9.8 x 5.2

  v = -38.96 m/s

ignoring the negative sign

magnitude of velocity is equal to 38.96 m/s

c) total distance travel by the stone

  vertical distance covered by the stone

 v² = u² + 2 g h

 0 = 12² - 2 x 9.8 x h

 h = 7.34 m

to reach the stone to the same level distance travel be doubled.

Total distance travel by the stone

H = h + h + 70

H = 7.34 x 2 + 70

H = 84.7 m.

8 0
3 years ago
How does radioactive decay work
Lesechka [4]

Answer:

Radioactive decay is the spontaneous breakdown of an atomic nucleus resulting in the release of energy and matter from the nucleus. Remember that a radioisotope has unstable nuclei that does not have enough binding energy to hold the nucleus together.

Explanation:

6 0
4 years ago
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