Natural gas is the answer.
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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.
Answer:
To create a second harmonic the rope must vibrate at the frequency of 3 Hz
Explanation:
First we find the fundamental frequency of the rope. The fundamental frequency is the frequency of the rope when it vibrates in only 1 loop. Therefore,
f₁ = v/2L
where,
v = speed of wave = 36 m/s
L = Length of rope = 12 m
f₁ = fundamental frequency
Therefore,
f₁ = (36 m/s)/2(12 m)
f₁ = 1.5 Hz
Now the frequency of nth harmonic is given in general, as:
fn = nf₁
where,
fn = frequency of nth harmonic
n = No. of Harmonic = 2
f₁ = fundamental frequency = 1.5 Hz
Therefore,
f₂ = (2)(1.5 Hz)
<u>f₂ = 3 Hz</u>
Answer:
128.9 N
Explanation:
The force exerted on the golf ball is equal to the rate of change of momentum of the ball, so we can write:

where
F is the force
is the change in momentum
is the time interval
The change in momentum can be written as

where
m = 0.04593 kg is the mass of the ball
u = 0 is the initial velocity of the ball
is the final velocity of the ball
Substituting into the original equation, we find the force exerted on the golf ball:
