Answer:
The velocity of the police car relative to earth is
Given:
time for flash generation of the inter galactic police car, t = 1.2 s
time between flashes as measured from earth, t' = 2.2 s
Solution:
Utilising Einstein's equation for time dilation to calculate the velocity of the police car, the equation is given by:
(1)
where, c = speed of light in vacuum =
re arranging eqn (1) for velocity, v:
(2)
Now, from eqn (2)
a)
b)
c)
Explanation:
a)
The energy stored in a capacitor is given by
where
C is the capacitance of the capacitor
V is the potential difference across the plates of the capacitor
For the capacitor in this problem, before insering the dielectric, we have:
is its capacitance
V = 22.0 V is the potential difference across it
Therefore, the initial energy stored in the capacitor is:
b)
After the dielectric is inserted into the plates, the capacitance of the capacitor changes according to:
where
k = 3.55 is the dielectric constant of the material
C is the initial capacitance of the capacitor
Therefore, the energy stored now in the capacitor is:
where:
is the initial capacitance
V = 22.0 V is the potential difference across the plate
Substituting, we find:
C)
The initial energy stored in the capacitor, before the dielectric is inserted, is
The final energy stored in the capacitor, after the dielectric is inserted, is
Therefore, the change in energy of the capacitor during the insertion is:
So, the energy of the capacitor has increased by
Answer:
Q= m ΔT and Q = m L
Explanation:
For this graph of temperature vs energy (heating) we must use two relations
* for when there is no change of state
Q= m ΔT
* for using there is change of state
Q = m L
the second expression is a consequence of the fact that all the energy supplied is used to change the state of the solid-liquid and liquid-gas system
the energy supplied is the sum of the energy in each interval
divide the system into intervals determined by the state change points
1) from T₀ = 70ºC to T_f = 327.4ºC, sample in solid-liquid state
c_e = 128 J / kg ºC
Q₁ = m c_e (T_f -To)
Q₁=1 128 (327.4 -70)
Q₁ = 3.29 10⁴ J
Q = Q₁ = 3.29 10⁴ J
2) when is it changing from solid to liquid
L = 2.45 10⁴ J / kg
Q2 = 1 2.45 10⁴
Q2 = 2.45 10⁴ J
Q = Q₁ + Q₂
Q = 5.74 10⁴ J
3) from to = 327.4ºC until T_f = 1725ºC, sample in liquid state
in the tables the specific heat of the solid and liquid state is the same
Q3 = m c_e (T_f -To)
Q3 = 1 128 (1725 -327.4)
Q3 = 1.79 10⁵ J
Q = Q₁ + Q₂ + Q₃
Q = (3.29 +2.45 + 17.9) 10⁴ J
Q = 23.64 10⁴ J
4) for when it is changing from the liquid state to the gaseous state
L_v = 8.70 10⁵ J / kg
Q₄ = m L_v
Q₄ = 1 8.70 10⁵
Q₄ = 8.70 10⁵ J
Q = Q₁ + Q₂ + Q₃ + Q₄
Q = (3.29 +5.74 + 17.9+ 87.0) 10⁴ J
Q = 110.64 10⁴ J
5) from To = 1725ºC to T_f = 2000ºC, sample in gaseous state
Q₅ = m c_e ΔT
Q₅ = 1 128 (2000 -1725)
Q₅ = 3.52 10⁴ J
Q = Q₁ + Q₂ + Q₃ + Q₄ + Q₅
Q = 114.16 104 J
the following table shows the points to be plotted
Energy (10⁴ J) Temperature (ºC)
0 70
3.29 327.4
5.74 327.4
23.64 1725
110.64 1725
114.16 2000
In the attachment we can see a graph of Temperature versus energy supplied
