Answer:

Explanation:
Consider two particles are initially at rest.
Therefore,
the kinetic energy of the particles is zero.
That initial K.E. = 0
The relative velocity with which both the particles are approaching each other is Δv and their reduced masses are

now, since both the masses have mass m
therefore,

= m/2
The final K.E. of the particles is

Distance between two particles is d and the gravitational potential energy between them is given by

By law of conservation of energy we have

Now plugging the values we get



This the required relation between G,m and d
Answer:
¨Facts you didn´t know¨ or ¨unknown facts¨
Explanation:
Answer:
U = – 0.12J
Explanation:
Given N = 10 turns, I = 5A, r = 5×10-²m
B^ = 0.05 T iˆ+ 0.3 T kˆ
Magnitude of the magnetic field vector B = √(0.05²+0.3²) = 0.304T
Area = πr² = π(5×10-²)² = 7.85×10-³m²
Magnetic moment μ = NIA
μ = 10×5×7.85×10-³ = 0.3925Am²
U = -μ•B = –0.3925×0.304 = –0.12J
The sign is negative because the magnetic moment is aligned with the magnetic field.
Answer:
<u>A:cool fluid sinks</u>
<u>B:warm fluid rises</u>
<u>C:convection current</u>
Explanation:
Just took the assessment!!