Because the Moon has a very small surface area compared to other spacial geo-bodies, it has cooled down much faster than Earth. Any water on the moon would freeze.
Answers:
a) 
b) 
c) 
Explanation:
<h3>a) Impulse delivered to the ball</h3>
According to the Impulse-Momentum theorem we have the following:
(1)
Where:
is the impulse
is the change in momentum
is the final momentum of the ball with mass
and final velocity (to the right) 
is the initial momentum of the ball with initial velocity (to the left) 
So:
(2)
(3)
(4)
(5)
<h3>b) Time </h3>
This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately
:
(6)
(7)
Where:
is the acceleration
is the length the ball was compressed
is the time
Finding
from (7):
(8)
(9)
(10)
Substituting (10) in (6):
(11)
Finding
:
(12)
<h3>c) Force applied to the ball by the bat </h3>
According to Newton's second law of motion, the force
is proportional to the variation of momentum
in time
:
(13)
(14)
Finally:

Answer:
To find the diameter of the wire, when the following are given:
Resistivity of the material (Rho), Current flowing in the conductor, I, Potential difference across the conductor ends, V, and length of the wire/conductor, L.
Using the ohm's law,
Resistance R = (rho*L)/A
R = V/I.
Crossectional area of the wire A = π*square of radius
Radius = sqrt(A/π)
Diameter = Radius/2 = [sqrt(A/π)]
Making A the subject of the formular
A = (rho* L* I)V.
From the result of A, Diameter can be determined using
Diameter = [sqrt(A/π)]/2. π is a constant with the value 22/7
Explanation:
Error and uncertainty can be measured varying the value of the parameters used and calculating different values of the diameters. Compare the values using standard deviation
Answer:
<em>20.08 Volts</em>
Explanation:
<u>Parallel Connection of Capacitors</u>
The voltage across any two elements connected in parallel is the same. If the elements are capacitors, then each voltage is


They are both the same after connecting them, thus

Or, equivalently

The total charge of both capacitors is

We can compute the total charge by using the initial conditions where both capacitors were disconnected:

Now we compute Q1 from the equation above

The final voltage of any of the capacitors is
