This player is initially at rest, then the Force Weight (W) and the Normal Force (N) have a same module. When applies a Extra Force (E) on the floor, this reacts with this increase, causing the player to be released upwards.
Using the Newton's Secound Law, we have:
Number 3If you notice any mistake in my english, please let me know, because i am not native.
Answer:
616.3 rad/s²
Explanation:
Given that
t= 1.46 s
Initial angular velocity ,ωi = 0 rad/s
Final angular velocity ωf= 27000 rev/min
Angular speed in the rad/s given as
Now by putting the values
ωf=900 rad/s
We know that (if acceleration is constant)
ωf=ωi + α t
α=Angular acceleration
900 = 0 + α x 1.46
Therefore the acceleration will be 616.3 rad/s²
First, we need to find the acceleration
F = ma
9850 N = (815 kg + 61 kg)a
a = 11.24
the read of the scale during the acceleration is :
mg + ma
= (61 kg x 9.8) + (61 kg x 11.24)
= 1283.4 N
hope this helps
Answer:
49.5 Hz.
Explanation:
From the question given above, the following data were obtained:
Period (T) = 0.0202 s
Frequency (f) =?
The frequency and period of a wave are related according to the following equation:
Frequency (f) = 1 / period (T)
f = 1/T
With the above formula, we can obtain the frequency of the wave as follow:
Period (T) = 0.0202 s
Frequency (f) =?
f = 1/T
f = 1/0.0202
f = 49.5 Hz
Therefore the frequency of the wave is 49.5 Hz.
