Answer:
Velocity = distance / time
V = 10/1
V = 10km/h
Answer:
1.7 m/s²
Explanation:
d = length of the ramp = 13.5 m
v₀ = initial speed of the skateboarder = 0 m/s
v = final speed of the skateboarder = 7.37 m/s
a = acceleration
Using the equation
v² = v₀² + 2 a d
7.37² = 0² + 2 a (13.5)
a = 2.01 m/s²
θ = angle of the incline relative to ground = 29.9
a' = Component of acceleration parallel to the ground
Component of acceleration parallel to the ground is given as
a' = a Cosθ
a' = 2.01 Cos29.9
a' = 1.7 m/s²
Answer:
The instantaneous velocity of the rocket the moment before it hits the ground is 50 m/s.
Explanation:
Given;
initial velocity of the rocket, u = 50 m/s
Determine the maximum height reached by the rocket.
at maximum height reached by the rocket, the final velocity, v = 0
v² = u² -2gh
0 = 50² - 2(9.8)h
19.6h = 2500
h = 2500 / 19.6
h = 127.55 m
At maximum height, the time to reach ground is given by;
h = ¹/₂gt²

Before the rocket hits the ground the final velocity will be maximum;
v = u + gt
v = 0 + 9.8 x 5.1
v = 50 m/s
Therefore, the instantaneous velocity of the rocket the moment before it hits the ground is 50 m/s.
We don't see any circuit diagrams.
This worries us for a few seconds, until we realize that we don't know anything about the experiment described in the problem either, so we don't have to worry about it at all.