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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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Let t be the time taken to stop.

ω = 0 rad/s

Use the first equation of motion for rotational motion

ω = ωo + α t

0 = 10.5 - 2 x t

t = 5.25 second

To learn more about angular acceleration, refer to:

brainly.com/question/21278452

#SPJ4

7 0
1 year ago
A world-class sprinter can burst out of the blocks to essentially top speed (of about 11.5 m/s) in the first 15.0 m of the race.
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Answer:

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Explanation:

Given:

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