The object is not moving, so the net force acting on it is zero.
The two given forces act in the same direction with a total magnitude of 2 N + 10 N = 12 N.
The third force acts counter to these with some magnitude <em>F</em> to give a net force of 0, so that by Newton's second law
<em>F</em> = 12 N
Answer:
The amount of charge on the plates is now equal to half its original value
Explanation:
From the question we are told that
The charge on the plate is Q
The initial separation is 
The new separation is 
Generally the capacitance of the capacitor is mathematically represented as
Generally the charge of the parallel plate capacitor is mathematically represented as
=> 
Here 
, A and V are constant , so looking at the question we see that Q varies inversely with d
So
Here K is a constant
so
=> 
=> ![Q_1 d_1 = Q_2 * [2 d_1]](https://tex.z-dn.net/?f=Q_1%20d_1%20%3D%20%20Q_2%20%2A%20%5B2%20d_1%5D)
=> 
So we see from the mathematically relation that the charge of the parallel plate capacitor will reduce by half of it original value
Answer:
The speed of man before he hits the ground is <u>23.35 m/s</u>
Explanation:
We know that:
Weight of Man - Force of Friction = Unbalanced Force
but, from Newton's 2nd Law of Motion:
unbalanced force = ma
Therefore,
W - F = ma
a = (W - F)/m
a = (mg - F)/m
where,
m = 81 kg
g = 9.8 m/s²
F = 103 N
a = [(81 kg)(9.8 m/s²) - 103 N]/81 kg
a = 8.52 m/s²
using 3rd equation of motion:
Vf² - Vi² = 2ah
here,
Vi = initial velocity = 0 m/s
Vf = Final Velocity before he hits ground = ?
Vf² - 0² = 2(8.52 m/s²)(32 m)
Vf = √545.28 m²/s²
<u>Vf = 23.35 m/s</u>
