Answer:
Part a)
V = 18.16 V
Part b)

Part c)
P = 672 Watt
Part d)
V = 5.84 V
Part e)

Explanation:
Part a)
When battery is in charging mode
then the potential difference at the terminal of the cell is more than its EMF and it is given as

here we have



now we have

Part b)
Rate of energy dissipation inside the battery is the energy across internal resistance
so it is given as



Part c)
Rate of energy conversion into EMF is given as



Now battery is giving current to other circuit so now it is discharging
now we have
Part d)



Part e)
now the rate of energy dissipation is given as



Answer:
1.36 x 10^-3 cm
Explanation:
Area = 50 ft^2 = 46451.5 cm^2
mass = 6 oz = 170.097 g
density = 2.70 g/cm^3
Let t be the thickness of foil in cm.
mass = volume x density
mass = area x thickness x density
170.097 = 46451.5 x t x 2.70
t = 1.36 x 10^-3 cm
Thus, the thickness of aluminium foil is 1.36 x 10^-3 cm.
Answer:
e. weight = mass x gravitational field strength
Answer:
e*P_s = 11 W
Explanation:
Given:
- e*P = 1.0 KW
- r_s = 9.5*r_e
- e is the efficiency of the panels
Find:
What power would the solar cell produce if the spacecraft were in orbit around Saturn
Solution:
- We use the relation between the intensity I and distance of light:
I_1 / I_2 = ( r_2 / r_1 ) ^2
- The intensity of sun light at Saturn's orbit can be expressed as:
I_s = I_e * ( r_e / r_s ) ^2
I_s = ( 1.0 KW / e*a) * ( 1 / 9.5 )^2
I_s = 11 W / e*a
- We know that P = I*a, hence we have:
P_s = I_s*a
P_s = 11 W / e
Hence, e*P_s = 11 W