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
The energy is 3.06 electronvolts, E = 3.06eV
1eV = 1.6 * 10^-19 J
3.06 eV = 3.06* 1.6 * 10^-19 J = 4.896 * 10^-19 J
Answer:
Explanation:
All the displacement will be converted into vector, considering east as x axis and north as y axis.
5.3 km north
D = 5.3 j
8.3 km at 50 degree north of east
D₁= 8.3 cos 50 i + 8.3 sin 50 j.
= 5.33 i + 6.36 j
Let D₂ be the displacement which when added to D₁ gives the required displacement D
D₁ + D₂ = D
5.33 i + 6.36 j + D₂ = 5.3 j
D₂ = 5.3 j - 5.33i - 6.36j
= - 5.33i - 1.06 j
magnitude of D₂
D₂²= 5.33² + 1.06²
D₂ = 5.43 km
Angle θ
Tanθ = 1.06 / 5.33
= 0.1988
θ =11.25 ° south of due west.
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