Answer:
The balloon would still move like a rocket
Explanation:
The principle of work of this system is the Newton's third law of motion, which states that:
"When an object A exerts a force on an object B (action), object B exerts an equal and opposite force (reaction) on object A"
In this problem, we can identify the balloon as object A and the air inside the balloon as object B. As the air goes out from the balloon, the balloon exerts a force (backward) on the air, and as a result of Newton's 3rd law, the air exerts an equal and opposite force (forward) on the balloon, making it moving forward.
This mechanism is not affected by the presence or absence of surrounding air: in fact, this mechanism also works in free space, where there is no air (and in fact, rockets also moves in space using this system, despite the absence of air).
It’s will be B because the circuit had a open or close so if that doesn’t work than it’s because it’s open
A moment causes a rotation about or axis. If the moment is to be taken about a point due to a force F, then in order for a moment to develop, the line of action cannot pass through that point...... the total moment was zero because the moment arm was zero as well
Answer:
(a) p = 3.4 kg-m/s (b) 37.78 N.
Explanation:
Mass of a basketball, m = 0.4 kg
Initial velocity of the ball, u = -5.7 m/s (as it comes down so it is negative)
It rebounds upward at a speed of 2.8 m/s (as it rebounds so positive)
(a) Change in momentum = final momentum - initial momentum
p = m(v-u)
p = 0.4 (2.8-(-5.7))
p = 3.4 kg-m/s
(b) Impulse = change in momentum
Ft = 3.4
We have, t = 0.09 s
Hence, this is the required solution.
Answer:
v = 50.5 m/s
Explanation:
F = (m)(^v/^t)
115N = (0.04551kg)(v/(0.020s))
2,526.917161 m/s² = v/(0.020s)
v = 50.53834322 m/s
v = 50.5 m/s
