-- The potential energy of a 12-lb bowling ball up on the shelf
doesn't have anything to do with the temperature of the ball or
the shelf.
-- The potential energy of a jar full of gas does depend on the
temperature of the gas. The warmer it is, the greater its pressure
is, and the more work it can do if you let it out through a little hole
in the jar. If it gets hot enough, it'll have enough potential energy
to blow the jar to smithereens.
Answer:
1) Q ’= 8 Q
, 2) q ’= 16 q
, 3) r ’= ¾ r
Explanation:
For this exercise we will use Coulomb's law
F = k q Q / r²
It asks us to calculate the change of any of the parameters so that the force is always F
Original values
q, Q, r
Scenario 1
q ’= 2q
r ’= 4r
F = k q ’Q’ / r’²
we substitute
F = k 2q Q ’/ (4r)²
F = k 2q Q '/ 16r²
we substitute the value of F
k q Q / r² = k q Q '/ 8r²
Q ’= 8 Q
Scenario 2
Q ’= Q
r ’= 4r
we substitute
F = k q ’Q / 16r²
k q Q / r² = k q’ Q / 16 r²
q ’= 16 q
Scenario 3
q ’= 3/2 q
Q ’= ⅜ Q
we substitute
k q Q r² = k (3/2 q) (⅜ Q) / r’²
r’² = 9/16 r²
r ’= ¾ r
Answer:0.153 Hz
Explanation: The relation between Time Period(T) and frequency(f) is given by T=1/f
Plug in the values and u arrive at the answer
Answer:
Explanation:
We shall apply law of conservation of momentum .
Momentum before collision = momentum after collision .
Momentum before collision = 400 kg m/s
Momentum after collision = 5 x v + 11 x 15
where v is velocity of A after the collision .
5 x v + 11 x 15 = 400
5 v = 400 - 165
5v = 235
v = 47 m /s .
