Explanation:
hope this helps you out if not im sorry
Answer:
μs = 0.75
μk = 0.58
Explanation:
From a force diagram:
(1)
(2)
When it starts slipping, friction force is the maximum and acceleration is 0. Replacing these conditions on (1):
Solving for μs:
μs = tan 36.7° = 0.75
When it moves at constant speed, friction force is kinetic friction and acceleration is 0. With these conditions the coefficient is:
μk = tan 30.1° = 0.58
Walking at a speed of 2.1 m/s, in the first 2 s John would have walked
(2.1 m/s) (2 s) = 4.2 m
Take this point in time to be the starting point. Then John's distance from the starting line at time <em>t</em> after the first 2 s is
<em>J(t)</em> = 4.2 m + (2.1 m/s) <em>t</em>
while Ryan's position is
<em>R(t)</em> = 100 m - (1.8 m/s) <em>t</em>
where Ryan's velocity is negative because he is moving in the opposite direction.
(b) Solve for the time when they meet. This happens when <em>J(t)</em> = <em>R(t)</em> :
4.2 m + (2.1 m/s) <em>t</em> = 100 m - (1.8 m/s) <em>t</em>
(2.1 m/s) <em>t</em> + (1.8 m/s) <em>t</em> = 100 m - 4.2 m
(3.9 m/s) <em>t</em> = 95.8 m
<em>t</em> = (95.8 m) / (3.9 m/s) ≈ 24.6 s
(a) Evaluate either <em>J(t)</em> or <em>R(t)</em> at the time from part (b).
<em>J</em> (24.6 s) = 4.2 m + (2.1 m/s) (24.6 s) ≈ 55.8 m
Answer:
40 N
Explanation:
We first need to calculate the acceleration of the tron ball.
Since acceleration, a = (v - u)/t where u = initial velocity of iron ball = 17m/s, v = final velocity of iron ball = 27m/s and t = time taken for the change in velocity = 5 s.
So, a = (v - u)/t
= (27 m/s - 17 m/s)/5 s
= 10 m/s ÷ 5 s
= 2 m/s²
We know force on iron ball, F = ma where m = mass of iron ball = 20 kg and a = acceleration = 2 m/s²
So, F = ma
= 20 kg × 2 m/s²
= 40 kgm/s²
= 40 N
So, the magnitude of the force on the iron ball is 40 N.
<span>The de-acceleration or negative acceleration of stopping is what damages bones. The ground is rigid and therefore the change in momentum when striking the ground will be large. On the trampoline, the elasticity of the material means that the momentum changes more slowly, resulting in smaller accelerations.</span>
