Ask Your Teacher It is found experimentally that the electric field in a certain region of Earth's atmosphere is directed vertic
ally down. At an altitude of 340 m the field has magnitude 60.0 N/C. At an altitude of 220 m, the magnitude is 100 N/C. Find the net amount of charge contained in a cube 120 m on edge, with horizontal faces at altitudes of 220 and 340 m. Neglect the curvature of Earth.
1 answer:
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Answer:
4800N
Explanation:
Lets assume,
Mass of first object = m₁
Mass of second object = m₂
Distance between the two objects = r
Thus the force between the two objects will be
where, G = Universal gravitational constant
Given, F = 2400N
New mass of second object = 2m₂
Now, the force will be
Thus, F₂ = 4800N
Answer:
(a) t = 1.14 s
(b) h = 0.82 m
(c) vf = 7.17 m/s
Explanation:
(b)
Considering the upward motion, we apply the third equation of motion:
where,
g = - 9.8 m/s² (-ve sign for upward motion)
h = max height reached = ?
vf = final speed = 0 m/s
vi = initial speed = 4 m/s
Therefore,
<u>h = 0.82 m</u>
Now, for the time in air during upward motion we use first equation of motion:
(c)
Now we will consider the downward motion and use the third equation of motion:
where,
h = total height = 0.82 m + 1.8 m = 2.62 m
vi = initial speed = 0 m/s
g = 9.8 m/s²
vf = final speed = ?
Therefore,
<u>vf = 7.17 m/s</u>
Now, for the time in air during downward motion we use the first equation of motion:
(a)
Total Time of Flight = t = t₁ + t₂
t = 0.41 s + 0.73 s
<u>t = 1.14 s</u>