Answer:
F = 15.47 N
Explanation:
Given that,
Q = 52 µC
q = 10 µC
d = 55 cm = 0.55 m
We need to find the magnitude of the electrostatic force on q. The formula for the electrostatic force is given by :
So, the magnitude of the electrostatic force is 15.47 N.
25,000 Feet = 7620m
PE = mgh where m is mass, g is gravity accel: 9.8 n h is height
= 90 x 9.8 x 7620
= 6720840J
= 6.72MJ
F = ma where m is mass, a is accel = gravity = 9.8
= 90 x 9.8
= 882N
Accel = gravity = 9.8m/s^2
KE = 1/2mv^2 where m is mass n v is vel
if no wind resistance, PE leaving airplane = KE at net
6720840 = 1/2 x 90 x v^2
v^2 = 149352
v = 386.5m/s
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the new momentum would be 40000 units as momentum = mass × velocity
