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
1)
first you find the maxium force that the car can produce.
f=ma
Fmax=(1100kg)(6m/s^2)
then use f = ma again to find the accel with the passengers
Fmax=(1100kg +1650kg)(a)
=> a = (1100kg)(6m/s^2)/( 1100kg +1650kg)
= 2.4 m/s^2
Answer:

Explanation:
= Mg ion = 
= F ion = 
q = Charge of electron = 
r = Distance between ions = 
k = Coulomb constant = 
Electrical force is given by
![F=-\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=-\dfrac{8.99\times 10^9\times 2\times 1.6\times 10^{-19}\times -1\times 1.6\times 10^{-19}}{[(0.072+0.133)\times 10^{-9}]^2}\\\Rightarrow F=1.09527\times 10^{-8}\ N](https://tex.z-dn.net/?f=F%3D-%5Cdfrac%7Bkq_1q_2%7D%7Br%5E2%7D%5C%5C%5CRightarrow%20F%3D-%5Cdfrac%7B8.99%5Ctimes%2010%5E9%5Ctimes%202%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D%5Ctimes%20-1%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D%7D%7B%5B%280.072%2B0.133%29%5Ctimes%2010%5E%7B-9%7D%5D%5E2%7D%5C%5C%5CRightarrow%20F%3D1.09527%5Ctimes%2010%5E%7B-8%7D%5C%20N)
The attractive force is 
Boyle’s law is P1V1 = P2V2. This shows that pressure and volume have an inverse relationship, so increase in pressure causes a decrease in volume. B is your answer
The answer is B hope this helps