Answer
given,
initial velocity of skateboard = 5.1 m/s
angle above the horizontal = 55°
height of the ramp = 1 m
a) maximum height of projectile
H = 0.889 m
the maximum height of the skateboard above the ground
= 1 + 0.889
= 1.889 m
b) time to reach the height
t = 0.426 s
horizontal distance = u cos θ × t
= 5.1 × cos 55° × 0.426
horizontal distance = 1.25 m
Answer:
True.
Explanation:
Defenintion of Muscular Endurance:
The ability of a muscle (or set of muscles) to perform a repeated action without tiring.
Let the mass of 2500 kg car be 
and it's velocity be 
and the mass of 1500 kg car be 
and it's velocity be 
.
After the bumping the mass be M and it's velocity be V.
By law of conservation of momentum we have
2500 * 5 + 1500 * 1=4000 * V
V = 14000/4000 = 7/2 = 3.5 m/s
So the velocity of the two-car train = 3.5 m/s
Pretty sure it’s Force*Distance*Cos(theta)
Frequency = 1/time period = 1/0.05 = 20s^-1.
