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victus00 [196]
3 years ago
5

It is found experimentally that the electric field in a certain region of Earth's atmosphere is directed vertically down. At an

altitude of 390 m the field has magnitude 60.0 N/C. At an altitude of 240 m, the magnitude is 100 N/C. Find the net amount of charge contained in a cube 150 m on edge, with horizontal faces at altitudes of 240 and 390 m. Neglect the curvature of Earth.
Physics
1 answer:
Maru [420]3 years ago
5 0

Answer:

q=7.965*10^-^6C

Explanation:

From the question we are told that

Altitude of  d_1 390m

MagnitudeM_1=60.0 N/C

Altitude of d_2=240 m

Magnitude is M_2= 100 N/C

Distance of cube d_c=150 m

Generally the flux \phi is mathematical given as

 \phi=60(150)^2cos180+100(150)^2*cos0

 \phi=-9*10^5

Generally Quantity of charge  q is mathematically given as

 q=\varepsilon _0 *\phi

 q=8.85*10^-^1^2 *9*10^5

 q=7.965*10^-^6C

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8 0
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Answer:

Explanation:

2as=vf^2-Vi^2

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What is the momentum of a photon having the same total energy as an electron with a kinetic energy of 100 keV?
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The momentum of the photon is 1.707 x 10⁻²² kg.m/s

Explanation:

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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

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