Answer:
23.52 m/s
Explanation:
The following data were obtained from the question:
Time taken (t) to reach the maximum height = 2.4 s
Acceleration due to gravity (g) = 9.8 m/s²
Initial velocity (u) =..?
At the maximum height, the final velocity (v) is zero. Thus, we can obtain how fast the rock (i.e initial velocity)
was thrown as follow:
v = u – gt (since the rock is going against gravity)
0 = u – (9.8 × 2.4)
0 = u – 23.52
Collect like terms
0 + 23.52 = u
u = 23.52 m/s
Therefore, the rock was thrown at a velocity of 23.52 m/s.
Explanation:
Artificial gravity can be created using a centripetal force. A centripetal force directed towards the center of the turn is required for any object to move in a circular path. In the context of a rotating space station it is the normal force provided by the spacecraft's hull that acts as centripetal force.
Hope it helps.
Answer:
I will answer in English.
Here we will use the relation
Velocity*time = distance
So:
a) velocity = 3m/s
time = 2s
Distance = 3m/s*2s = 6m
b) velocity = 2m/s
time = 3.5s
Distance = 2m/s*3.5s = 7m
c) velocity = 10m/s
time = 0.5s
Distance = 10m/s*0.5s = 5m
d) velocity = 4m/s
time = 2.5s
Distance = 4m/s*2.5s = 9m
e) velocity = 1.5m/s
time = 5s
Distance = 1.5m/s*5s = 7.5m
I’m pretty sure it does most of the time ig
The solution for this problem is:
500 revolution per
minute = 8.33rev /s = 2π*8.33 rad /s = 52.36 rad /s
Angular velocity ω = 2π N
Angular acceleration α= (ω2 - ω1) /t
ω2 = 0
α = - ω1/t = -2π N /t
N = 500 rpm = 8.33 r p s.
α = -2π 8.33 /2.6 =- 20 rad/s^2
