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
Alpha emission is the process results in a change in mass number. Option B is correct.
<h3>What is mass number?</h3>
The total number of protons and neutrons in an atomic nucleus is known as the mass number, often known as the atomic mass number or nucleon number.
It's about the same as the atom's atomic mass, expressed in atomic mass units.
The alpha particle is a helium nucleus with two protons and two neutrons in an alpha decay or alpha emission. The number of protons and neutrons is reduced by two as a result of this action.
The quantity of protons and neutrons is affected by gamma emission descent. Also, while electron capture has no effect on the number of neutrons, it does raise the 1 also number of protons by one.
Alpha emission is the process results in a change in mass number.
Hence option B is correct.
To learn more about the mass number, refer:
brainly.com/question/4408975
#SPJ1
Answer:

Explanation:
consider the mass of each train car be m
m₁ = m₂ = m₃ = m
speed of the three identical train
u₁ = u₂ = u₃ = 1.8 m/s
m₄ = m u₄ = 4.5 m/s
m₅ = m u₅ = 0 (initial velocity )
final velocity
v₁ = v₂ = v₃ = v₄ = v₅ = v
using conservation of momentum
m₁u₁ + m₂u₂ + m₃u₃ + m₄u₄ + m₅u₅ = m₁v₁ + m₂v₂ + m₃v₃ + m₄v₄ + m₅v₅
m (1.8 + 1.8 + 1.8 +4.5) = 5 m v


Answer:
<h2>
6.36 cm</h2>
Explanation:
Using the formula to first get the image distance
1/f = 1/u+1/v
f = focal length of the lens
u = object distance
v = image distance
Given f = 16.0 cm, u = 24.8 cm
1/v = 1/16 - 1/24.8
1/v = 0.0625-0.04032
1/v = 0.02218
v = 1/0.02218
v = 45.09 cm
To get the image height, we will us the magnification formula.
Mag = v/u = Hi/H
Hi = image height = ?
H = object height = 3.50 cm
45.09/24.8 = Hi/3.50
Hi = (45.09*3.50)/24.8
Hi = 6.36 cm
The image height is 6.36 cm
I think the correct answer from the choices listed above is option B. The very high voltage needed to create a spark across the spark plug is produced at the transformer's secondary winding. <span>The secondary coil is engulfed by a powerful and changing magnetic field. This field induces a current in the coils -- a very high-voltage current.</span>