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nika2105
15 days ago
9

What are the two principal parts of an AC induction motor? 

Physics
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If a metal wire is 4m long and a force of 5000n causes it to stretch by 1mm, what is the strain?
serg [3582]

Answer:

2.5\cdot 10^{-4}

Explanation:

Strain is defined as the ratio of an object's dimensional change when subjected to a force:

S=\frac{\Delta L}{L_0}

where

\Delta L indicates the alteration in length of the object

L_0 signifies the object's initial length

In this case, we have L_0 = 4 m and \Delta L=1 mm=0.001 m, hence the strain is

S=\frac{\Delta L}{L_0}=\frac{0.001 m}{4 m}=2.5\cdot 10^{-4}


5 0
2 months ago
On a caterpillars map all distances are marked in kilometers . The caterpillars map shows the distance between two milkweed plan
ValentinkaMS [3465]

Answer:

The equivalent distance in kilometers is 4012 ×10^{-6} km.

Explanation:

It's known that 1 millimeter converts to 10^{-3} meters. Then, 1 meter converts to 10^{-3} kilometers. Therefore, the conversion for 1 millimeter to kilometers can be stated as

1 mm = 10^{-3} m

1 m = 10^{-3} km

Thus, 1 mm = 10^{-3}×10^{-3} km = 10^{-6} km.

Given the distance of 4012 mm, the corresponding distance in kilometers will be

4012 mm = 4012 ×10^{-6} km.

The distance therefore is 4012 ×10^{-6} km.

5 0
4 months ago
A spinning wheel is slowed down by a brake, giving it a constant angular acceleration of 25.60 rad/s2. During a 4.20-s time inte
Yuliya22 [3333]

We will use the equations of rotational kinematics,

\theta =\theta _{0} + \omega_{0} t+ \frac{1}{2}\alpha t^2             (A)

\omega^2= \omega^2_{0} +2\alpha\theta                                     (B)                                          

Here, \theta and \theta _{0} denote the final and initial angular displacements, respectively, whereas \omega and \omega_{0} represent final and initial angular velocities, and \alpha is the angular acceleration.

We are provided with \alpha = - 25.60 \ rad/s^2, \theta = 62.4 \ rad and t = 4.20 \ s.

By substituting these values into equation (A), we have

62.4 \ rad = 0 + \omega_{0} 4.20 \ s + \frac{1}{2} (- 25.60 \ rad/s^2) ( 4.20)^2 \\\\ \omega_{0} = \frac{220.5+ 62.4 }{4.20} =67.4 \ rad/s

Now, using equation (B),

\omega^2=(67.4 \ rad/s)^2 + 2 (- 25.60 \ rad/s^2)62.4 \ rad \\\\\ \omega = 36.7 \ rad/s

This indicates that the wheel's angular speed at the 4.20-second mark is 36.7 rad/s.

4 0
3 months ago
A person who climbs up something (e.g., a hill, a ladder, the stairs) from the ground gains potential energy. a person's weight
kicyunya [3294]

The following values have been provided:

weight w = 240 lb = 1,067.52 N

energy E = 3,000 J

 

The equation for potential energy is:

E = w h

where h indicates the height that the person needs to ascend, therefore:

h = 3000 / 1067.52

h = 2.81 m

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<span>Thus, he must ascend 2.81 meters</span>

3 0
3 months ago
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