That depends on the wavelength itself, like what type of wave it is and where on the spectrometer it is
We Know,
K.E. = 1/2 mv²
480 = 1/2 (m)(8)²
m = 960/64
m = 15 Kg
So, the mass of the object is 15 Kg
Answer:
The current is 
Explanation:
From the question we are told that
The radius is 
The current density is 
The distance we are considering is 
Generally current density is mathematically represented as

Where A is the cross-sectional area represented as

=> 
=> 
Now the change in current per unit length is mathematically evaluated as

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows


![I = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ] | \left 0.001585} \atop 0}} \right.](https://tex.z-dn.net/?f=I%20%20%3D%202%5Cpi%20%2A%289.0%2A10%5E%7B6%7D%29%20%5B%5Cfrac%7Br%5E4%7D%7B4%7D%20%5D%20%20%7C%20%5Cleft%20%20%20%200.001585%7D%20%5Catop%200%7D%7D%20%5Cright.)
![I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ]](https://tex.z-dn.net/?f=I%20%20%3D%202%5Cpi%20%2A%289.0%2A10%5E%7B6%7D%29%20%5B%20%5Cfrac%7B0.001585%5E4%7D%7B4%7D%20%5D)
substituting values
![I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ]](https://tex.z-dn.net/?f=I%20%20%3D%202%20%2A%20%203.142%20%20%2A%20%209.00%20%2A10%5E6%20%2A%20%20%20%5B%20%5Cfrac%7B0.001585%5E4%7D%7B4%7D%20%5D)

If you double the velocity of an object, its momentum doubles. but for the same increase in velocity, the kinetic energy increases 4 times..