The bouncing person because the bounce helped him survive
(a) 328.6 kg m/s
The linear impulse experienced by the passenger in the car is equal to the change in momentum of the passenger:

where
m = 62.0 kg is the mass of the passenger
is the change in velocity of the car (and the passenger), which is

So, the linear impulse experienced by the passenger is

(b) 404.7 N
The linear impulse experienced by the passenger is also equal to the product between the average force and the time interval:

where in this case
is the linear impulse
is the time during which the force is applied
Solving the equation for F, we find the magnitude of the average force experienced by the passenger:

Average Velocity=Total Distance/Total Time


Wavelength = (speed) / (frequency)
= (3 x 10⁸ m/s) / (1 x 10⁸ /s) = 3 meters
Gamma ray<span> (also called </span>gamma radiation<span>), denoted by the lower-case Greek letter </span>gamma,<span> is penetrating </span>electromagnetic radiation<span> of a kind arising from the </span>radioactive decay<span> of </span>atomic nuclei<span>. It consists of </span>photons<span> in the highest observed range of </span>photon energy<span>. </span>