To solve this problem it is necessary to apply the concepts related to Young's Module and its respective mathematical and modular definitions. In other words, Young's Module can be expressed as

Where,
F = Force/Weight
A = Area
= Compression
= Original Length
According to the values given we have to




Replacing this values at our previous equation we have,



Therefore the Weight of the object is 3.82kN
0.2 is the value of coefficient of friction (k)
F=kN
F=horizontal force
n=Normal Force
k=coefficient of friction
k=F/N
k=200/1000
k=0.2
The ratio of the normal force pushing two surfaces together to the frictional force preventing motion between them is known as the friction coefficient. Usually, the Greek letter mu is used to indicate it .N is the normal force, and F is the frictional force, hence F = N/N.
Due to the fact that both F and N are measured in units of force, the coefficient of friction has no dimensions (such as newtons or pounds). The coefficient of friction can have a variety of values for both static and dynamic friction. Static friction occurs when an object encounters friction that resists any applied force, keeping the object at rest until the static frictional force is released. In kinetic friction, the frictional force resists the motion of the object.
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Answer:
1.04%
Explanation:
Given that,
The power of drill = 3.5 kW = 3500 W
Transferred kinetic energy = 5000 kJ during 15 seconds of use.
We need to find the percentage efficiency of the drill. It can be given by :

Where
Po and Pi are output and input powers.

So,

So, the percentage efficiency of the drill is 1.04%.
Answer:
The magnitude of the electric field at the center of the circle is 85 N/C
The correct option is b. 85 N/C
Explanation:
Please see the attachments below
Answer:
(a) 
(b) 
Explanation:
We can derive the initial speed of the rock from the equation of the speed in function of the time:

Using the given values for the speed at time t=1.7s, we get:

In words, the speed of the rock at launch is 34m/s (a).
Next, we use this to calculate the speed at t=4.9s:

This means that the speed of the rock at 4.9s after the launch is 14m/s (b), and the negative sign means that it is moving downwards.