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).
The time of motion of the track star is determined as 0.837 s.
<h3>Time of motion of the track star</h3>
The time of motion of the track star is calculated as follows;
T = (2u sinθ)/g
where;
- T is time of motion
- g is acceleration due to gravity
- θ is angle of projection
T = (2 x 12 x sin20)/9.8
T = 0.837 s
Learn more about time of motion here: brainly.com/question/2364404
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V = 8 * 10^2 km/h = 800km/h
S= 1,8* 10^3 km = 1800km
t = ?
v = S/t
t = S/v
t = 1800km/ 800km/h
t ≈ 2,25h (135min)
Answer:
a) In order to catch the ball at the level at which it is thrown in the direction of motion.
b)Speed of the receiver will be 7.52m/s
Explanation:
Calculating range,R= Vo^2Sin2theta/g
R= (20^2×Sin(2×30)/9.8 = 35.35m
Let receiver be(R-20) = 35.35-20= 15.35m
The horizontal component of the ball is:
Vox= Vocostheta= 20× cos30°
Vox= 17.32m/s
Time taken to coverR=35.35m with 17.32m/s will be:
t=R/Vox= 35.35/17.32
t= 2.04seconds
b)Speed required to cover 15.35m at 2.04seconds
Vxreciever= d/t = 15.35/2.04 = 7.52m/s
