Answer:
The energy of the capacitors connected in parallel is 0.27 J
Given:
C = 
C' = 
Potential difference, V = 300 V
Solution:
Now, we know that the equivalent capacitance of the two parallel connected capacitors is given by:
The energy of the capacitor, E is given by;
Answers are:
(1) KE = 1 kg m^2/s^2
(2) KE = 2 kg m^2/s^2
(3) KE = 3 kg m^2/s^2
(4) KE = 4 kg m^2/s^2
Explanation:
(1) Given mass = 0.125 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.125 * (16)
KE = 1 kg m^2/s^2
(2) Given mass = 0.250 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.250 * (16)
KE = 2 kg m^2/s^2
(3) Given mass = 0.375 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.375 * (16)
KE = 3 kg m^2/s^2
(4) Given mass = 0.500 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.5 * (16)
KE = 4 kg m^2/s^2
Answer:
= 0.331 J / g ° C
Explanation:
We have a calorimetry exercise where all the heat yielded by one of the components is absorbed by the other.
Heat ceded Qh = m1 ce1 (
-
)
Heat absorbed Qc = m2 ce2 ( - T₀)
Body 1 is metal and body 2 is water
. Where m are the masses of the two bodies, ce their specific heat and T the temperatures
Qh = Qc
m₁ (- ) = m₂ 
( - T₀)
we clear the specific heat of the metal
= m₂ ( - T₀) / (m₁ (-))
= 50.00 4.184 (20.15 -10.79) / (75.00 (99.0-20.15))
= 209.2 (9.36) / (75 78.85)
= 1958.11 / 5913.75
= 0.331 J / g ° C
Distance from the reference point
We calculate current from the formula:
, where q is a electric charge transferred over time t
Time should be converted to seconds:
1h 15 min= 75min= 4500s
I=
Result is in unit-Ampere
