The emf is induced in the wire will be 1.56 ×10 ⁻³ V. The induced emf is the product of the magnetic field,velocity and length of the wire.
<h3>What is induced emf?</h3>
Emf is the production of a potential difference in a coil as a result of changes in the magnetic flux passing through it.
When the flux coupling with a conductor or coil changes, electromotive Force, or EMF, is said to be induced.
The given data in the problem is;
B is the magnitude of the magnetic field,= 5.0 ×10⁻⁵ T
V(velocity)=125 M/SEC
L(length)=25 cm=0.25 m
The maximum emf is found as;
E=VBLsin90°
E=125 × 5.0 × 10⁻⁵ ×0.25
E=1.56 ×10 ⁻³ V
Hence, the emf is induced in the wire will be 1.56 ×10 ⁻³ V
To learn more about the induced emf, refer to the link;
brainly.com/question/16764848
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Assume that the small-massed particle is 
and the heavier mass particle is 
.
Now, by momentum conservation and energy conservation:
Now, there are 2 solutions but, one of them is useless to this question's main point so I excluded that point. Ask me in the comments if you want the excluded solution too.
So now, we see that 
and 
. So therefore, the smaller mass recoils out.
Hope this helps you!
Bye!
The answer is c because the farther apart they are the greater there gravity is
It makes no difference. The momentum of either car goes to zero in both cases.
