The momentum of an object is equivalent to the product of the object's mass and velocity. Computing the momentum for each ball:
A- 15 * 0.7 = 10.5
B- 5.5 * 1.2 = 6.6
C- 5.0 * 2.5 = 12.5
D- 1.5 * 5.0 = 7.5
Therefore, ball C has the greatest momentum.
Both valves are closed during the power stroke.
While the fuel is burning in the cylinder, you want
all the force of the expanding gases to push the
piston down ... you don't want any of the gases
or their pressure escaping.
If either of the valves was open, even just a crack,
then part of the gases would go blooey out the valve,
and some pressure would be lost that's supposed to be
pushing the piston.
Jennifer
Explanation:
she has more mass which means she is using more force
Answer:
Part a)

Part b)

Part c)

Part d)

Part e)

Part f)

Explanation:
As we know that catapult is projected with speed 19.9 m/s
so here we have


similarly we have


Part a)
Horizontal displacement in 1.03 s



Part b)
Vertical direction we have
![y = v_y t - \frac{1]{2}gt^2](https://tex.z-dn.net/?f=y%20%3D%20v_y%20t%20-%20%5Cfrac%7B1%5D%7B2%7Dgt%5E2)


Part c)
Horizontal displacement in 1.71 s



Part d)
Vertical direction we have
![y = v_y t - \frac{1]{2}gt^2](https://tex.z-dn.net/?f=y%20%3D%20v_y%20t%20-%20%5Cfrac%7B1%5D%7B2%7Dgt%5E2)


Part e)
Horizontal displacement in 5.44 s



Part f)
Vertical direction we have
![y = v_y t - \frac{1]{2}gt^2](https://tex.z-dn.net/?f=y%20%3D%20v_y%20t%20-%20%5Cfrac%7B1%5D%7B2%7Dgt%5E2)


Answer:
The amount of work the factory worker must to stop the rolling ramp is 294 joules
Explanation:
The object rolling down the frictionless ramp has the following parameters;
The mass of the object = 10 kg
The height from which the object is rolled = 3 meters
The work done by the factory worker to stop the rolling ramp = The initial potential energy, P.E., of the ramp
Where;
The potential energy P.E. = m × g × h
m = The mass of the ramp = 10 kg
g = The acceleration due to gravity = 9.8 m/s²
h = The height from which the object rolls down = 3 m
Therefore, we have;
P.E. = 10 kg × 9.8 m/s² × 3 m = 294 Joules
The work done by the factory worker to stop the rolling ramp = P.E. = 294 joules