Answer: B
Action and reaction forces act on the same object and go to the same direction
<h2>The work done = - 2 x 10⁴ J</h2>
Explanation:
In the first case , the volume is kept constant and pressure varies .
In isothermal process , the work done
W₁ = V x ΔP
here V is the volume of gas and ΔP is the change in pressure
Thus W₁ = 0
Because there is no change in volume , therefore displacement is zero .
In second case pressure is constant , but volume changes
Thus W₂ = P x ΔV
here P is the pressure and ΔV is the change in volume
Therefore W₂ = 4 x 10⁵ x 5 x 10⁻² = 2 x 10⁴ J
The total work done W = - 2 x 10⁴ J
Because the work done in compression is negative .
Answer:
- <u>The water ballon that was thrown straight down at 2.00 m/s hits the ground first, 0.19 s before the other ballon.</u>
Explanation:
The motions of the two water ballons are ruled by the kinematic equations:
We are only interested in the vertical motion, so that equation is all what you need.
<u>1. Water ballon is thrown horizontally at sped 2.00 m/s.</u>
The time the ballon takes to hit the ground is independent of the horizontal speed.
Since 2.00 m/s is a horizontal speed, you take the initial vertical speed equal to 0.
Then:

<u>2. Water ballon thrown straight down at 2.00 m/s</u>
Now the initial vertical speed is 2.00 m/s down. So, the equation is:

To solve the equation you can use the quadratic formula.

You get two times. One of the times is negative, thus it does not have physical meaning.
<u>3. Conclusion:</u>
The water ballon that was thrown straight down at 2.00 m/s hits the ground first by 1.11 s - 0.92s = 0.19 s.
The question is incomplete. The complete question is :
Assume that the energy lost was entirely due to friction and that the total length of the PVC pipe is 1 meter. Use this length to compute the average force of friction (for this calculation, you may neglect uncertainties).
Mass of the ball : 16.3 g
Predicted range : 0.3503 m
Actual range : 1.09 m
Solution :
Given that :
The predicted range is 0.3503 m
Time of the fall is :

...........(i)
...........(ii)
Dividing the equation (ii) by (i)

∴ 
Now loss of energy = change in the kinetic energy
![$W=\frac{1}{2} m [v_0^2-v_1^2]$](https://tex.z-dn.net/?f=%24W%3D%5Cfrac%7B1%7D%7B2%7D%20m%20%5Bv_0%5E2-v_1%5E2%5D%24)
![$W=\frac{1}{2} \times (16.3 \times 10^{-3}) \times [v_0^2-\left(\frac{v_0}{3.11}\right)^2]$](https://tex.z-dn.net/?f=%24W%3D%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20%2816.3%20%5Ctimes%2010%5E%7B-3%7D%29%20%5Ctimes%20%5Bv_0%5E2-%5Cleft%28%5Cfrac%7Bv_0%7D%7B3.11%7D%5Cright%29%5E2%5D%24)

If f is average friction force, then
(f)(L) = W
(f) (1) = 
(f) = 