Answer:
<h2>50m/s^2</h2>
Explanation:
Step one:
given data
initial velocity u= 0m/s since the ball is at rest
time of contact t= 0.3s
final velocity v=15m/s
Required
acceleration a
from the first law of motion
v=u+at
substitute our given data
15=0+a*0.3
15=0.3a
divide both sides by 0.3
a=15/0.3
a=50m/s
<u>The average acceleration is 50m/s^2</u>
Answer:
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I assume the 100 N force is a pulling force directed up the incline.
The net forces on the block acting parallel and perpendicular to the incline are
∑ F[para] = 100 N - F[friction] = 0
∑ F[perp] = F[normal] - mg cos(30°) = 0
The friction in this case is the maximum static friction - the block is held at rest by static friction, and a minimum 100 N force is required to get the block to start sliding up the incline.
Then
F[friction] = 100 N
F[normal] = mg cos(30°) = (10 kg) (9.8 m/s²) cos(30°) ≈ 84.9 N
If µ is the coefficient of static friction, then
F[friction] = µ F[normal]
⇒ µ = (100 N) / (84.9 N) ≈ 1.2
Answer:
13,750 N
Yes
Explanation:
Given:
v₀ = 90 km/h = 25 m/s
v = 0 m/s
t = 4 s
Find: a and Δx
a = Δv / Δt
a = (0 m/s − 25 m/s) / (4 s)
a = -6.25 m/s²
F = ma
F = (2200 kg) (-6.25 m/s²)
F = -13,750 N
Δx = ½ (v + v₀) t
Δx = ½ (0 m/s + 25 m/s) (4 s)
Δx = 50 m
the cycle of processes by which water circulates between the earth's oceans, atmosphere and land involving precipitation as rain and snow drainage in streams and rivers and return to the atmosphere by evaporation and transpiration.
Hope this gives you a little bit more information!