The continent of Antartica is located at the bottom of the world. the South Pole is at its center. Antarctica is the coldest and windiest place on earth. It is covered with ice up to 3 miles thick. Very few plants and animals can survive here, but penguins, fish, and seals live on the coast and in the seas. No people live on Antarctica permanently, but scientists and tourists visit.
Answer:
A.
Explanation:
an atom that has most of its mass is electrons.
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)
Answer:
![E=1.62\times 10^{14}\ J](https://tex.z-dn.net/?f=E%3D1.62%5Ctimes%2010%5E%7B14%7D%5C%20J)
Explanation:
Given that,
The mass of the paperclip, m = 1.8 g = 0.0018 kg
We need to find the energy obtained. The relation between mass and energy is given by :
![E=mc^2](https://tex.z-dn.net/?f=E%3Dmc%5E2)
Where
c is the speed of light
So,
![E=0.0018\times (3\times 10^8)^2\\\\E=1.62\times 10^{14}\ J](https://tex.z-dn.net/?f=E%3D0.0018%5Ctimes%20%283%5Ctimes%2010%5E8%29%5E2%5C%5C%5C%5CE%3D1.62%5Ctimes%2010%5E%7B14%7D%5C%20J)
So, the energy obtained is
.