Answer:
The magnitude of the uniform magnetic field exerting this torque on the loop is 1.67 T
Explanation:
Given;
radius of the wire, r = 0.45 m
current on the loop, I = 2.4 A
angle of inclination, θ = 36⁰
torque on the coil, τ = 1.5 N.m
The torque on the coil is given by;
τ = NIBAsinθ
where;
B is the magnetic field
Area of the loop is given by;
A = πr² = π(0.45)² = 0.636 m
τ = NIBAsinθ
1.5 = (1 x 2.4 x 0.636 x sin36)B
1.5 = 0.8972B
B = 1.5 / 0.8972
B = 1.67 T
Therefore, the magnitude of the uniform magnetic field exerting this torque on the loop is 1.67 T
The one that does that is the photoelectric effect. This allows the sun to hit the surface and produces enough energy for electrons to be knocked off the atom and allows a current to move.
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!
Kinetic energy = 1/2 × m × v^2
16 = 1/2 × m × 2^2
16 = 1/2 × m × 4
16 = 2 × m
16/2 = m
8 = m
so the mass is 8 kg
