Answer:
M1 V1 = M1 V2 + M2 V3 conservation of momentum
V2 = (M1 V1 - M2 V3) / M1 where V2 = speed of M1 after impact
V2 = (3 * 9 - 1.5 * 5) / 9 = (27 - 7.5) / 9 = 2.17 m/s
Note: All speeds are in the same direction and have the same sign
T² caries directly as R³ .
This is Kepler's 3rd law of planetary motion .
Answer:
I think C
Explanation:
Since the bus is moving away from John.
{C - V}.
Answer:
<h2>2.4</h2>
Explanation:
<h2><em>Hope it help mark as Brainlist</em></h2>
Complete Question
A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.
Answer:
The value ![v_2 = 4 \sqrt{10} \ m/s](https://tex.z-dn.net/?f=v_2%20%20%3D%20%204%20%5Csqrt%7B10%7D%20%5C%20%20m%2Fs)
Explanation:
From the question we are told that
The charge on the first sphere is ![q_1 = 2\mu C = 2*10^{-6} \ C](https://tex.z-dn.net/?f=q_1%20%20%3D%20%202%5Cmu%20C%20%20%3D%20%202%2A10%5E%7B-6%7D%20%5C%20%20C)
The charge on the second sphere is ![q_2 = 8 \mu C = 8*10^{-6} \ C](https://tex.z-dn.net/?f=q_2%20%3D%20%208%20%5Cmu%20C%20%3D%208%2A10%5E%7B-6%7D%20%5C%20%20C)
The mass of the second charge is ![m = 1.50 \ g = 1.50 *10^{-3} \ kg](https://tex.z-dn.net/?f=m%20%20%3D%20%201.50%20%5C%20%20g%20%20%3D%20%201.50%20%2A10%5E%7B-3%7D%20%5C%20kg)
The distance apart is ![d = 0.4 \ m](https://tex.z-dn.net/?f=d%20%3D%20%200.4%20%5C%20%20m)
The speed of the second sphere is ![v_1 = 20 \ ms^{-1}](https://tex.z-dn.net/?f=v_1%20%20%3D%20%2020%20%5C%20%20ms%5E%7B-1%7D)
Generally the total energy possessed by when
and
are separated by
is mathematically represented
![Q = KE + U](https://tex.z-dn.net/?f=Q%20%3D%20%20KE%20%2B%20U)
Here KE is the kinetic energy which is mathematically represented as
![KE = \frac{1 }{2} m (v_1)^2](https://tex.z-dn.net/?f=KE%20%20%3D%20%20%5Cfrac%7B1%20%7D%7B2%7D%20%20m%20%28v_1%29%5E2)
substituting value
![KE = \frac{1 }{2} * ( 1.50 *10^{-3}) (20 )^2](https://tex.z-dn.net/?f=KE%20%20%3D%20%20%5Cfrac%7B1%20%7D%7B2%7D%20%20%2A%20%28%201.50%20%2A10%5E%7B-3%7D%29%20%2820%20%29%5E2)
![KE = 0.3 \ J](https://tex.z-dn.net/?f=KE%20%20%3D%20%200.3%20%5C%20%20J)
And U is the potential energy which is mathematically represented as
![U = \frac{k * q_1 * q_2 }{d }](https://tex.z-dn.net/?f=U%20%20%3D%20%20%5Cfrac%7Bk%20%2A%20%20q_1%20%2A%20%20q_2%20%20%7D%7Bd%20%7D)
substituting values
![U = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.8 }](https://tex.z-dn.net/?f=U%20%20%3D%20%20%5Cfrac%7B9%2A10%5E9%20%2A%20%202%2A10%5E%7B-6%7D%20%2A%208%2A10%5E%7B-6%7D%20%20%7D%7B0.8%20%7D)
![U = 0.18 \ J](https://tex.z-dn.net/?f=U%20%20%3D%20%200.18%20%5C%20%20J)
So
![Q = 0.3 + 0.18](https://tex.z-dn.net/?f=Q%20%3D%20%200.3%20%2B%20%200.18)
![Q = 0.48 \ J](https://tex.z-dn.net/?f=Q%20%3D%20%200.48%20%5C%20%20J)
Generally the total energy possessed by when
and
are separated by
is mathematically represented
![Q_f = KE_f + U_f](https://tex.z-dn.net/?f=Q_f%20%3D%20%20KE_f%20%2B%20U_f)
Here
is the kinetic energy which is mathematically represented as
![KE_f = \frac{1 }{2} m (v_2^2](https://tex.z-dn.net/?f=KE_f%20%20%3D%20%20%5Cfrac%7B1%20%7D%7B2%7D%20%20m%20%28v_2%5E2)
substituting value
![KE_f = \frac{1 }{2} * ( 1.50 *10^{-3}) (v_2 )^2](https://tex.z-dn.net/?f=KE_f%20%20%3D%20%20%5Cfrac%7B1%20%7D%7B2%7D%20%20%2A%20%28%201.50%20%2A10%5E%7B-3%7D%29%20%28v_2%20%29%5E2)
![KE_f = 7.50 *10^{ -4} (v_2 )^2](https://tex.z-dn.net/?f=KE_f%20%20%3D%20%207.50%20%2A10%5E%7B%20-4%7D%20%28v_2%20%29%5E2)
And
is the potential energy which is mathematically represented as
![U_f = \frac{k * q_1 * q_2 }{d }](https://tex.z-dn.net/?f=U_f%20%20%3D%20%20%5Cfrac%7Bk%20%2A%20%20q_1%20%2A%20%20q_2%20%20%7D%7Bd%20%7D)
substituting values
![U_f = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.4 }](https://tex.z-dn.net/?f=U_f%20%20%3D%20%20%5Cfrac%7B9%2A10%5E9%20%2A%20%202%2A10%5E%7B-6%7D%20%2A%208%2A10%5E%7B-6%7D%20%20%7D%7B0.4%20%7D)
![U_f = 0.36 \ J](https://tex.z-dn.net/?f=U_f%20%20%3D%20%200.36%20%5C%20%20J)
From the law of energy conservation
![Q = Q_f](https://tex.z-dn.net/?f=Q%20%3D%20%20Q_f)
So
![0.48 = 0.36 +(7.50 *10^{-4} v_2^2)](https://tex.z-dn.net/?f=0.48%20%3D%20%200.36%20%2B%287.50%20%2A10%5E%7B-4%7D%20v_2%5E2%29)
![v_2 = 4 \sqrt{10} \ m/s](https://tex.z-dn.net/?f=v_2%20%20%3D%20%204%20%5Csqrt%7B10%7D%20%5C%20%20m%2Fs)