Answer:
Answer:
4 ms
Explanation:
initial velocity, u = 75 m/s
final velocity, v = 0
distance, s = 15 cm = 0.15 m
Let the acceleration is a and the time taken is t.
Use third equation of motion
v² = u² + 2 a s
0 = 75 x 75 - 2 a x 0.15
a = - 18750 m/s^2
Use first equation of motion
v = u + at
0 = 75 - 18750 x t
t = 4 x 10^-3 s
t = 4 ms
thus, the time taken is 4 ms.
Explanation:
<span>EP (potential energy) = mgy -> (59)(9.8)(-5) = -2,891
EP + EK (kinetic energy) = 0; but rearranging it for EK makes it EK = -EP, such that EK = 2891 when plugged in.
EK = 0.5mv^2, but can also be v = sqrt(2EK/m).
Plugging that in for sqrt((2 * 2891)/59), we get 9.9 m/s^2 with respect to significant figures.</span>
Answer:
The velocity of the cart at the bottom of the ramp is 1.81m/s, and the acceleration would be 3.30m/s^2.
Explanation:
Assuming the initial velocity to be zero, we can obtain the velocity at the bottom of the ramp using the kinematics equations:
Dividing the second equation by the first one, we obtain:
And, since , then:
It means that the velocity at the bottom of the ramp is 1.81m/s.
We could use this data, plus any of the two initial equations, to determine the acceleration:
So the acceleration is 3.30m/s^2.