Answer:
a) 0.504 kJ
b) 67.7 kJ
c) 14.9 kJ
d) 25.5%
Explanation:
<u>note:</u>
<u>solution is attached in word form due to error in mathematical equation. futhermore i also attach Screenshot of solution in word because to different version of MS Office please find the attachment</u>
Answer:
770.1 m
Explanation:
From the question given above, the following data were obtained:
Velocity (v) = 45.3 m/s
Time (t) = 17 s
Displacement (d) =?
Velocity is defined according to the following formula:
Velocity = Displacement /Time
With the above formula, we can obtain the displacement of the ball as follow:
Velocity = Displacement /Time
45.3 = Displacement / 17
Cross multiply
Displacement = 45.3 × 17
Displacement = 770.1 m
Therefore the displacement of the ball is 770.1 m
Answer:
The work done is ![32.2kJ](https://tex.z-dn.net/?f=32.2kJ)
Explanation:
Work is defined as the product of force and distance moved in the direction of application of force.
![W= F*S](https://tex.z-dn.net/?f=W%3D%20F%2AS)
Given Data
distance S= 230m
force F= 140 N
Mass of car m= 1140 kg
Applying the formula for work done we have
![W= 140*230\\W= 32200 J\\W= 32.2kJ](https://tex.z-dn.net/?f=W%3D%20%20140%2A230%5C%5CW%3D%2032200%20J%5C%5CW%3D%2032.2kJ)
The work done by pushing the car for a distance of
is ![32.2kJ](https://tex.z-dn.net/?f=32.2kJ)
I would use a resistor rated for 1 W or more. Not less.
As per Einstein's theory of relativity we know that when an object will move with the speed comparable to the speed of light then the length of the object will be different from its length at rest position
This is also known as length contraction theory
As we know here that
![L = L_0\sqrt{1 - \frac{v^2}{c^2}}](https://tex.z-dn.net/?f=L%20%3D%20L_0%5Csqrt%7B1%20-%20%5Cfrac%7Bv%5E2%7D%7Bc%5E2%7D%7D)
so here we know that
v = 0.95 c
so from above equation we will have
![L = L_0\sqrt{1 - 0.95^2}](https://tex.z-dn.net/?f=L%20%3D%20L_0%5Csqrt%7B1%20-%200.95%5E2%7D)
![L = 0.31 L_0](https://tex.z-dn.net/?f=L%20%3D%200.31%20L_0)
so here the length will be SHORTER