Answer:
The momentum of the photon is 1.707 x 10⁻²² kg.m/s
Explanation:
Given;
kinetic of electron, K.E = 100 keV = 100,000 eV = 100,000 x 1.6 x 10⁻¹⁹ J = 1.6 x 10⁻¹⁴ J
Kinetic energy is given as;
K.E = ¹/₂mv²
where;
v is speed of the electron
![K.E = \frac{1}{2}mv^2\\\\mv^2 = 2K.E \\\\v^2 = \frac{2K.E}{m} \\\\v = \sqrt{\frac{2K.E}{m}} \\\\but \ momentum ,P = mv\\\\(v)m = (\sqrt{\frac{2K.E}{m}})m\\\\P_{photon} = (\sqrt{\frac{2K.E}{m_e}})m_e\\\\P_{photon} = (\sqrt{\frac{2\times 1.6\times 10^{-14}}{9.11\times10^{-31}}})(9.11\times 10^{-31})\\\\P_{photon} = 1.707 \times 10^{-22} \ kg.m/s](https://tex.z-dn.net/?f=K.E%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2%5C%5C%5C%5Cmv%5E2%20%3D%202K.E%20%5C%5C%5C%5Cv%5E2%20%3D%20%5Cfrac%7B2K.E%7D%7Bm%7D%20%5C%5C%5C%5Cv%20%3D%20%5Csqrt%7B%5Cfrac%7B2K.E%7D%7Bm%7D%7D%20%5C%5C%5C%5Cbut%20%5C%20momentum%20%2CP%20%3D%20mv%5C%5C%5C%5C%28v%29m%20%3D%20%28%5Csqrt%7B%5Cfrac%7B2K.E%7D%7Bm%7D%7D%29m%5C%5C%5C%5CP_%7Bphoton%7D%20%3D%20%20%28%5Csqrt%7B%5Cfrac%7B2K.E%7D%7Bm_e%7D%7D%29m_e%5C%5C%5C%5CP_%7Bphoton%7D%20%3D%20%20%28%5Csqrt%7B%5Cfrac%7B2%5Ctimes%201.6%5Ctimes%2010%5E%7B-14%7D%7D%7B9.11%5Ctimes10%5E%7B-31%7D%7D%7D%29%289.11%5Ctimes%2010%5E%7B-31%7D%29%5C%5C%5C%5CP_%7Bphoton%7D%20%3D%201.707%20%5Ctimes%2010%5E%7B-22%7D%20%5C%20kg.m%2Fs)
Therefore, the momentum of the photon is 1.707 x 10⁻²² kg.m/s
Answer:
b
Explanation:
The space shuttle, in circular orbit around the Earth, collides with a small asteroid which ends up in the shuttle's storage bay.
This form of collision is called inelastic collision. And inelastic collision momentum is conserved but the kinetic energy is not conserved. Hence the correct option is b. only momentum is conserved.
14 ms is required to reach the potential of 1500 V.
<u>Explanation:</u>
The current is measured as the amount of charge traveling per unit time. So the charge of electrons required for each current is determined as the product of current with time.
![Charge = Current \times Time](https://tex.z-dn.net/?f=Charge%20%3D%20Current%20%5Ctimes%20Time)
As two different current is passing at two different times, the net charge will be the different in current. So,
![\text { Charge }=(1.0000020-1.0000000) \times t=2 \times 10^{-6} \times t](https://tex.z-dn.net/?f=%5Ctext%20%7B%20Charge%20%7D%3D%281.0000020-1.0000000%29%20%5Ctimes%20t%3D2%20%5Ctimes%2010%5E%7B-6%7D%20%5Ctimes%20t)
The electric voltage on the surface of cylinder can be obtained as the ratio of charge to the radius of the cylinder.
![V=\frac{k q}{R}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7Bk%20q%7D%7BR%7D)
Here
, q is the charge and R is the radius. As
and R =17 cm = 0.17 m, then the voltage will be
![V=\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B9%20%5Ctimes%2010%5E%7B9%7D%20%5Ctimes%202%20%5Ctimes%2010%5E%7B-6%7D%20%5Ctimes%20t%7D%7B0.17%7D)
The time is required to find to reach the voltage of 1500 V, so
![1500 =\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}](https://tex.z-dn.net/?f=1500%20%3D%5Cfrac%7B9%20%5Ctimes%2010%5E%7B9%7D%20%5Ctimes%202%20%5Ctimes%2010%5E%7B-6%7D%20%5Ctimes%20t%7D%7B0.17%7D)
![\begin{aligned}&t=\frac{1500 \times 0.17}{\left(9 \times 10^{9} \times 2 \times 10^{-6}\right)}\\&t=14.1666 \times 10^{-3} s=14\ \mathrm{ms}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%26t%3D%5Cfrac%7B1500%20%5Ctimes%200.17%7D%7B%5Cleft%289%20%5Ctimes%2010%5E%7B9%7D%20%5Ctimes%202%20%5Ctimes%2010%5E%7B-6%7D%5Cright%29%7D%5C%5C%26t%3D14.1666%20%5Ctimes%2010%5E%7B-3%7D%20s%3D14%5C%20%5Cmathrm%7Bms%7D%5Cend%7Baligned%7D)
So, 14 ms is required to reach the potential of 1500 V.
A rotating disc supplied with constant power where the relationship of the angular velocity of the disc and the number of rotations made by the disc is governed by Newton's second law for rotation. This law is specially made for rotating bodies which is extracted from Newton's second law of motion.
New Moon
Waxing Crescent
First Quarter
<span>Waxing Gibbous
</span>Full Moon
<span>Waning Gibbous
</span>Last Quarter
Waning Crescent
;)