Joe and Jim start at rest at the starting line of a race track. Both runners accelerate at the same rate, but Joe accelerates for twice as long as Jim. Describe the final speed of Joe compared to Jim.

Vf₂ = 2 Vf₁

It shows that final speed of Joe is twice the final speed of Jim.

Explanation:

First, we analyze the final speed of Jim by using first equation of motion:

Vf₁ = Vi + at

where,

Vf₁ = final speed of Jim

Vi = initial speed of Jim = 0 m/s

a = acceleration of Jim

t = time of acceleration for Jim

Therefore,

Vf₁ = at   equation (1)

Now, we see the final speed of Joe. For Joe the parameters will become:

Vf = Vf₂

Vi = 0 m/s

a = a

t = 2t

Therefore,

Vf₂ = 2at  

using equation (1):

Vf₂ = 2 Vf₁

It shows that final speed of Joe is twice the final speed of Jim.

3.2x10^7

you'd use the equation that establishes a relationship between  energy  and  wavelength, which looks like this

e=h⋅cλ, where

λ  - the wavelength of the photon; c  - the speed of light, usually given as  3⋅108m/s; h  -  planck's constant, equal to  6.626⋅10−34j s

plug the value you were given for the energy of the photon in the above equation to get the wavelength

e=h⋅cλ⇒λ=h⋅ce

λ=6.626⋅10−34js⋅3⋅108ms−1.257⋅10−24j=1.581⋅10−1m