An x-ray photon is scattered by an originally stationary electron. how does the frequency of the scattered photon compare relati
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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
The new velocity after 4 s is 40 m/s
The height of the spaceship above the ground after 5 seconds is 1,127.5 m
The given parameters for the first question;
- initial velocity of the car, u = 76 m/s
- acceleration of the car, a = - 9 m/s²
- time of motion, t = 4 s
The new velocity after 4 s is calculated as;
v = u + at
v = 76 + (-9)(4)
v = 76 - 36
v = 40 m/s
(5)
The given parameters;
- height above the ground, h = 500 m
- velocity of spaceship, u = 150 m/s
- time of motion, t = 5
The height of the spaceship above the ground after 5 seconds is calculated as;
Learn more here: brainly.com/question/24527971