Answer:
1027 N/C
3.42 x 10⁻⁶ T
Explanation:
I = Intensity of electromagnetic field = 1400 W/m²
E₀ = Maximum value of electric field
Intensity of electromagnetic field is given as
I = (0.5) ε₀ E₀² c
1400 = (0.5) (8.85 x 10⁻¹²) (3 x 10⁸) E₀²
E₀ = 1027 N/C
B₀ = maximum value of magnetic field
using the equation
E₀ = B₀ c
1027 = B₀ (3 x 10⁸)
B₀ = 3.42 x 10⁻⁶ T
Answer:
+131Joules
Explanation:
Energy can be expressed using below expresion.
ΔE = (q + w).........eqn(1)
q will be + be if heat is gained hence, q= 240 J
work "w" will be - ve if work is done by the system, hence w= -109 J
Then substitute into eqn(1)
Change in Internal energy=
= (240 -109 )
= +131J
Radius of nuclei is given by formula

now we can say volume of the nuclei is given as

now the density is given as
density = mass / volume
mass of nuclei = mass of neutron + mass of protons



Now density is given as

here we know that
= 1.2 fm


So from above we can say that density of all nuclei is almost same.
Answer:
(a) 43.2 kC
(b) 0.012V kWh
(c) 0.108V cents
Explanation:
<u>Given:</u>
- i = current flow = 3 A
- t = time interval for which the current flow =

- V = terminal voltage of the battery
- R = rate of energy = 9 cents/kWh
<u>Assume:</u>
- Q = charge transported as a result of charging
- E = energy expended
- C = cost of charging
Part (a):
We know that the charge flow rate is the electric current flow through a wire.

Hence, 43.2 kC of charge is transported as a result of charging.
Part (b):
We know the electrical energy dissipated due to current flow across a voltage drop for a time interval is given by:

Hence, 0.012V kWh is expended in charging the battery.
Part (c):
We know that the energy cost is equal to the product of energy expended and the rate of energy.

Hence, 0.108V cents is the charging cost of the battery.