Answer:
Drag or air resistance
Explanation:
The force of friction caused by a moving fluid is called drag.  When that fluid is air, it's also known as air resistance.
 
        
             
        
        
        
Kinetic energy lost in collision is 10 J.
<u>Explanation:</u>
Given,
Mass,  = 4 kg
 = 4 kg
Speed,  = 5 m/s
 = 5 m/s
 = 1 kg
 = 1 kg
 = 0
 = 0
Speed after collision = 4 m/s
Kinetic energy lost, K×E = ?
During collision, momentum is conserved.
Before collision, the kinetic energy is

By plugging in the values we get,

K×E = 50 J
Therefore, kinetic energy before collision is 50 J
Kinetic energy after collision:


Since,
Initial Kinetic energy = Final kinetic energy
50 J = 40 J + K×E(lost)
K×E(lost) = 50 J - 40 J
K×E(lost) = 10 J
Therefore, kinetic energy lost in collision is 10 J.
 
        
             
        
        
        
Answer:
Distance covered by B is 4 times distance covered by A
Explanation:
For an object in free fall starting from rest, the distance covered by the object in a time t is

where
s is the distance covered
g is the acceleration due to gravity
t is the time elapsed
In this problem:
- Object A falls through a distance  during a time t, so the distance covered by object A is
 during a time t, so the distance covered by object A is

- Object B falls through a distance  during a time 2t, so the distance covered by object B is
 during a time 2t, so the distance covered by object B is

So, the distance covered by object B is 4 times the distance covered by object A.
 
        
             
        
        
        
The type of energy that depends on position is called
kinetic energy
        
             
        
        
        
Answer:
Approximately  . (Assuming that the drag on this ball is negligible, and that
. (Assuming that the drag on this ball is negligible, and that  .)
.)
Explanation:
Assume that the drag (air friction) on this ball is negligible. Motion of this ball during the descent:
- Horizontal: no acceleration, velocity is constant (at  is constant throughout the descent.) is constant throughout the descent.)
- Vertical: constant downward acceleration at  , starting at , starting at . .
The horizontal velocity of this ball is constant during the descent. The horizontal distance that the ball has travelled during the descent is also given:  . Combine these two quantities to find the duration of this descent:
. Combine these two quantities to find the duration of this descent:
 .
.
In other words, the ball in this question start at a vertical velocity of  , accelerated downwards at
, accelerated downwards at  , and reached the ground after
, and reached the ground after  .
.
Apply the SUVAT equation  to find the vertical displacement of this ball.
 to find the vertical displacement of this ball.
![\begin{aligned}& x(\text{vertical}) \\[0.5em] &= -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t\\[0.5em] &= - \frac{1}{2} \times 10\; \rm m \cdot s^{-2} \times (7.5\; \rm s)^{2} \\ & \quad \quad + 0\; \rm m \cdot s^{-1} \times 7.5\; s \\[0.5em] &= -281.25\; \rm m\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%26%20x%28%5Ctext%7Bvertical%7D%29%20%5C%5C%5B0.5em%5D%20%26%3D%20-%5Cfrac%7B1%7D%7B2%7D%5C%2C%20g%20%5Ccdot%20t%5E%7B2%7D%20%2B%20v_0%5Ccdot%20t%5C%5C%5B0.5em%5D%20%26%3D%20-%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%2010%5C%3B%20%5Crm%20m%20%5Ccdot%20s%5E%7B-2%7D%20%5Ctimes%20%287.5%5C%3B%20%5Crm%20s%29%5E%7B2%7D%20%5C%5C%20%26%20%5Cquad%20%5Cquad%20%2B%200%5C%3B%20%5Crm%20m%20%5Ccdot%20s%5E%7B-1%7D%20%5Ctimes%207.5%5C%3B%20s%20%5C%5C%5B0.5em%5D%20%26%3D%20-281.25%5C%3B%20%5Crm%20m%5Cend%7Baligned%7D) .
.
In other words, the ball is  below where it was before the descent (hence the negative sign in front of the number.) The height of this cliff would be
 below where it was before the descent (hence the negative sign in front of the number.) The height of this cliff would be  .
.