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
Answer:
t = 0.657 s
Explanation:
First, let's use the appropiate equations to solve this:
V = √T/u
This expression gives us a relation between speed of a disturbance and the properties of the material, in this case, the rope.
Where:
V: Speed of the disturbance
T: Tension of the rope
u: linear density of the rope.
The density of the rope can be calculated using the following expression:
u = M/L
Where:
M: mass of the rope
L: Length of the rope.
We already have the mass and length, which is the distance of the rope with the supports. Replacing the data we have:
u = 2.31 / 10.4 = 0.222 kg/m
Now, replacing in the first equation:
V = √55.7/0.222 = √250.9
V = 15.84 m/s
Finally the time can be calculated with the following expression:
V = L/t ----> t = L/V
Replacing:
t = 10.4 / 15.84
t = 0.657 s
Answer:
Time taken = 8.25 second
Explanation:
Given:
Force = 4000 N
Force = ma
4,000 = (1100)(a)
Acceleration = 3.6363 m/s²
v = u + at
0 = 30 + (3.6363)t
Time taken = 8.25 second
