Coulomb's Law
Given:
F = 3.0 x 10^-3 Newton
d = 6.0 x 10^2 meters
Q1 = 3.3x 10^-8 Coulombs
k = 9.0 x 10^9 Newton*m^2/Coulombs^2
Required:
Q2 =?
Formula:
F = k • Q1 • Q2 / d²
Solution:
So, to solve for Q2
Q2 = F • d²/ k • Q1
Q2 = (3.0 x 10^-3 Newton) • (6.0 x 10^2 m)² / (9.0 x 10^9
Newton*m²/Coulombs²) • (3.3x 10^-8 Coulombs)
Q2 = (3.0 x 10^-3 Newton) • (360 000 m²) / (297 Newton*m²/Coulombs)
Q2 = 1080 Newton*m²/ (297 Newton*m²/Coulombs)
Then, take the reciprocal of the denominator and start
multiplying
Q2 = 1080 • 1 Coulombs/297
Q2 = 1080 Coulombs / 297
Q2 = 3.63636363636 Coulombs
Q2 = 3.64 Coulumbs
He doesn't lose weight he stays the same weight it's just gravity that changes
In this case, volume of the can remains constant. The relationship between pressure and temperature at constant volume is given by:
P/T = Constant
Then

Where
P1 = 40 psi
P2 = ?
T1 = 60°F ≈ 289 K
T2 = 90°F ≈ 305 K (note, 363 K is not right)
Substituting;
The car's rate of acceleration : a = 2.04 m/s²
<h3>Further explanation</h3>
Given
speed = 110 km/hr
time = 15 s
Required
The acceleration
Solution
110 km/hr⇒30.56 m/s
Acceleration is the change in velocity over time
a = Δv : Δt
Input the value :
a = 30.56 m/s : 15 s
a = 2.04 m/s²
Answer:x=23.4 cm
Explanation:
Given
mass of block 
inclination 
coefficient of static friction 
coefficient of kinetic friction 
distance traveled 
spring constant 
work done by gravity+work done by friction=Energy stored in Spring





