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
Piper rockelle and I just got off the phone number
Answer: a) 3.85 days
b) 10.54 days
Explanation:-
Expression for rate law for first order kinetics is given by:
where,
k = rate constant = ?
t = time taken for decomposition = 3 days
a = let initial amount of the reactant = 100 g
a - x = amount left after decay process = 
First we have to calculate the rate constant, we use the formula :
Now put all the given values in above equation, we get
a) Half-life of radon-222:
Thus half-life of radon-222 is 3.85 days.
b) Time taken for the sample to decay to 15% of its original amount:
where,
k = rate constant = 
t = time taken for decomposition = ?
a = let initial amount of the reactant = 100 g
a - x = amount left after decay process = 
Thus it will take 10.54 days for the sample to decay to 15% of its original amount.
Answer:
The voltage will be 0.0125V
Explanation:
See the picture attached
Answer:
The work done on the suitcase is, W = 1691 J
Explanation:
Given data,
The force on the suitcase is, F = 89 N
The distance Russell dragged the suitcase, S = 19 m
The work done on the suitcase by Russell is equal to the work done on the suitcase to overcome the friction
The work done on the suitcase by Russell is given by the formula
W = F · S
Substituting the given values,
W = 89 N x 19 m
W = 1691 J
Hence, the work done on the suitcase is, W = 1691 J
