Answer:
Moment of the force is 20 N-m.
Explanation:
Given:
Force exerted by the person is, 
Distance of application of force from the point about which moment is needed is, 
Now, we know that, moment of a force 'F' about a point at a perpendicular distance of 'd' from the same point is given as the product of the force and the perpendicular distance.
Therefore, the moment of the force about the end of the claw hammer is given as:
Hence, the moment of the force exerted by the person about the end of the claw hammer is 20 N-m.
the SL unit of acceleration is the meter per second squared
Answer:
f = 409 Hz
Explanation:
We have,
Length of the open organ pipe, l = 0.29 m
Frequency of vibration of second overtone, 
It is required to find the fundamental frequency of the pipe. For the open organ pipe, the frequency of second overtone is given by :
v is speed of sound
Let f is the fundamental frequency. It is given by :
The relation between f and f₂ can be written as :
So, the fundamental frequency of the pipe is 409 Hz.
Answer:
A) 31 kJ
B) 1.92 KJ
C) 40 , 2.48
Explanation:
weight of person ( m ) = 79 kg
height of jump ( h ) = 0.510 m
Compression of joint material ( d ) = 1.30 cm ≈ 0.013 m
A) calculate the force
Fd = mgh
F = mgh / d
W = mg
F(net) = W + F = mg ( 1 + 
= 79 * 9.81 ( 1 + (0.51 / 0.013) )
= 774.99 ( 40.231 ) ≈ 31 KJ
B) calculate the force when the stopping distance = 0.345 m
d = 0.345 m
Fd = mgh hence F = mgh / d
F(net) = W + F = mg ( 1 +
= 79 * 9.81 ( 1 + (0.51 / 0.345) )
= 774.99 ( 2.478 ) = 1.92 KJ
C) Ratio of force in part a with weight of person
= 31000 / ( 79 * 9.81 ) = 31000 / 774.99 = 40
Ratio of force in part b with weight of person
= 1920 / 774.99 = 2.48
