The velocity of B after elastic collision is 3.45m/s
This type of collision is an elastic collision and we can use a formula to solve this problem.
<h3>Elastic Collision</h3>

The data given are;
- m1 = 281kg
- u1 = 2.82m/s
- m2 = 209kg
- u2 = -1.72m/s
- v1 = ?
Let's substitute the values into the equation.

From the calculation above, the final velocity of the car B after elastic collision is 3.45m/s.
Learn more about elastic collision here;
brainly.com/question/7694106
<h2>v = 2.5J/c</h2>
Explanation:
<h2>Given:</h2>
<h3>Energy = E = 20J</h3>
<h3>Charge = Q = 8C</h3>
<h2>Require :</h2>
___________
<h3>Potential Difference = V = ?</h3>
<h2>Formula :</h2>
____________
<h2>V =

</h2>
<h2>Solution:</h2><h3>_________</h3>
<h2>V =

</h2>
<h3>V = 2.5V</h3>
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.
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<h2>

</h2>
Lets say that:
Eric is E
Sheena is S
Yael is Y
We know that S+2=Y, E= 1/S and that E+S+Y=52. So, we can insert S+2=Y and E= 1/S into our other equation, which would bring us to (1/2S)+(S+2)+S=52. This is equal to 1/2S+S+2+S=52. We then use algebra to get the answer:
Given:
S+2=Y
E=1/S
E+Y+S=52
Equation:
1/2S+S+2+S=52
1/2S+S+S=52
2.5S=50
S=50/2.5
S=20
And since Yael has 2 more points then Yael then we do 20+2 and get 22.
Hope this helps! <3
When person is observing destructive interference at 0.20 m distance from the equidistant position then we can say that path difference must be equal to half of the wavelength
now we will have

now we know that
y = 0.20 m
d = 2.4 m
L = 10 m
now here we have


now frequency of wave is given as


To solve this problem we will use Snell's law. This law is used to calculate the angle of refraction of light by crossing the separation surface between two means of propagation of light (or any electromagnetic wave) with a different index of refraction.

Where,
= Index of refraction for each material
= Angle of incidence and refraction
A ray of light propagating in a medium with index of refraction
hitting an angle
on the surface of a medium of index
with
can be fully reflected inside the medium of highest refractive index:

Refractive index of rarer medium
Refractive index of denser medium
Replacing we have,



the critical angle of this fiber at the glass-plastic interface is 60.07°