Answer:
![|\Delta v |=\sqrt{\frac{4Gm}{d} }](https://tex.z-dn.net/?f=%7C%5CDelta%20v%20%7C%3D%5Csqrt%7B%5Cfrac%7B4Gm%7D%7Bd%7D%20%7D)
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
![\mu= \frac{m_1m_2}{m_1+m_2}](https://tex.z-dn.net/?f=%5Cmu%3D%20%5Cfrac%7Bm_1m_2%7D%7Bm_1%2Bm_2%7D)
now, since both the masses have mass m
therefore,
![\mu= \frac{m^2}{2m}](https://tex.z-dn.net/?f=%5Cmu%3D%20%5Cfrac%7Bm%5E2%7D%7B2m%7D)
= m/2
The final K.E. of the particles is
![KE_{final}=\frac{1}{2}\times \mu\times \Delta v^2](https://tex.z-dn.net/?f=KE_%7Bfinal%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cmu%5Ctimes%20%5CDelta%20v%5E2)
Distance between two particles is d and the gravitational potential energy between them is given by
![PE_{Gravitational}= \frac{Gmm}{d}](https://tex.z-dn.net/?f=PE_%7BGravitational%7D%3D%20%5Cfrac%7BGmm%7D%7Bd%7D)
By law of conservation of energy we have
![KE_{initial}+KE_{final}= PE_{gravitaional}](https://tex.z-dn.net/?f=KE_%7Binitial%7D%2BKE_%7Bfinal%7D%3D%20PE_%7Bgravitaional%7D)
Now plugging the values we get
![0+\frac{1}{2}\frac{m}{2}\Delta v^2= -\frac{Gmm}{d}](https://tex.z-dn.net/?f=0%2B%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bm%7D%7B2%7D%5CDelta%20v%5E2%3D%20-%5Cfrac%7BGmm%7D%7Bd%7D)
![|\Delta v |=\sqrt{\frac{4Gm}{d} }](https://tex.z-dn.net/?f=%7C%5CDelta%20v%20%7C%3D%5Csqrt%7B%5Cfrac%7B4Gm%7D%7Bd%7D%20%7D)
![=\sqrt{\frac{Gm}{d} }](https://tex.z-dn.net/?f=%3D%5Csqrt%7B%5Cfrac%7BGm%7D%7Bd%7D%20%7D)
This the required relation between G,m and d
<span>Every planet with an atmosphere has clouds. That includes the moon.</span>
Answer: A
it's a sign that there might be something causing too much
current to flow in the circuit
Explanation: When a fuse blows, one of the causes may be short circuit. Also, it's a sign that there might be something causing too much
current to flow in the circuit leading to overloading. That is, drawing to much of power that can cause the metal inside the fuse to melt.
Option A is the best answer.
INCREASE in temperature of the material practically increase the energy of the particles. which increases their motion due to increase in energy . thus when the temperature is decreased the energy level decreases which causes the particle's motion to slow down.. the motion of the particle is highly reduced when the temperature is lowered
<span>The 2nd truck was overloaded with a load of 16833 kg instead of the permissible load of 8000 kg.
The key here is the conservation of momentum.
For the first truck, the momentum is
0(5100 + 4300)
The second truck has a starting momentum of
60(5100 + x)
And finally, after the collision, the momentum of the whole system is
42(5100 + 4300 + 5100 + x)
So let's set the equations for before and after the collision equal to each other.
0(5100 + 4300) + 60(5100 + x) = 42(5100 + 4300 + 5100 + x)
And solve for x, first by adding the constant terms
0(5100 + 4300) + 60(5100 + x) = 42(14500 + x)
Getting rid of the zero term
60(5100 + x) = 42(14500 + x)
Distribute the 60 and the 42.
60*5100 + 60x = 42*14500 + 42x
306000 + 60x = 609000 + 42x
Subtract 42x from both sides
306000 + 18x = 609000
Subtract 306000 from both sides
18x = 303000
And divide both sides by 18
x = 16833.33
So we have the 2nd truck with a load of 16833.33 kg, which is well over it's maximum permissible load of 8000 kg. Let's verify the results by plugging that mass into the before and after collision momentums.
60(5100 + 16833.33) = 60(21933.33) = 1316000
42(5100 + 4300 + 5100 + 16833.33) = 42(31333.33) = 1316000
They match. The 2nd truck was definitely over loaded.</span>