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).
Answer:
網站有中、英文版本，也有繁、簡體版，可通過每頁左上角的連結隨時網站有中、英文版本，也有繁、簡體版，可通過每頁左上角的連結隨時
Explanation:
Answer:
The velocity of the particle from T = 0 s to T = 4 s is;
0.5 m/s
Explanation:
The given parameters from the graph are;
The initial displacement (covered) at time, t₁ = 0 s is x₁ = 1 m
The displacement covered at time, t₂ = 4 s is x₂ = 3 m
The graph of distance to time, from time t = 0 to time t = 4 is a straight line graph, with the velocity given by the rate of change of the displacement to the time which is dx/dt which is also the slope of the graph given as follows;
The velocity of the particle from t = 0 s to t = 4 s = 1/2 m/s = 0.5 m/s.
