HELP NEEDED PLEASE
1 answer:
You might be interested in
Answer:
Force exerted, F = 1.5 N
Explanation:
It is given that, a boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s.
i.e. u = 0
v = 30 m/s
Time taken, t = 0.06 s
Mass of the paper, m = 0.003 kg
We need to find the force the boxer exert on it. The force can be calculated using second law of motion as :
F = 1.5 N
So, the force the boxer exert on the paper is 1.5 N. Hence, this is the required solution.
The fundamental frequency of the tube is 0.240 m long, by taking air temperature to be C is 367.42 Hz.
A standing wave is basically a superposition of two waves propagating opposite to each other having equal amplitude. This is the propagation in a tube.
The fundamental frequency in the tube is given by
where,
Since, T=37+273 K = 310 K
v = 331 m/s
Using this, we get:
Hence, the fundamental frequency is 367.42 Hz.
To learn more about Attention here:
#SPJ4