Explanation:
Given:
m = 1.673 × 10^-27 kg
Q = q = 1.602 × 10^-19 C
r = 0.75 nm
= 0.75 × 10^-9 m
A.
Energy, U = (kQq)/r
Ut = 1/2 mv^2 + 1/2 mv^2
1.673 × 10^-27 × v^2 = (8.99 × 10^9 × (1.602 × 10^-19)^2)/0.75 × 10^-9
v = 1.356 × 10^4 m/s
B.
F = (kQq)/r^2
F = m × a
1.673 × 10^-27 × a = ((8.99 × 10^9 × (1.602 × 10-19)^2)/(0.075 × 10^-9)^2
a = 2.45 × 10^17 m/s^2.
The andwer of tye question is 3O2
Vertical forces:
There is a force of 579N acting upward, and a force of 579N
acting downward.
The vertical forces are balanced ... they add up to zero ...
so there's no vertical acceleration.
Not up, not down.
Horizontal forces:
There is a force of 487N acting to the left, and a force of 632N
acting to the right.
The net horizontal force is
(487-left + 632-right) - (632-right - 487-right) = 145N to the right.
The net force on the car is all to the right.
The car accelerates to the right.
We use a fundamental kinematic equation as follows:
V = Vo + g*t.
<span>Tr = (V-Vo)/g = (0-10)/-10 = 1 s. = </span><span>time to reach max. height </span>
<span>Tf = Tr = 1 s. = Fall time or time to fall back to edge of bldg. </span>
<span>3-Tr-Tf = 3-1-1 = 1 s. Below edge of bldg. </span>
<span>d = Vo*t + 0.5g*t^2. </span>
<span>d = 10*1 + 5*1^2 = 15 m. <---- OPTION C</span>
