Answer:
Current, I = 1000 A
Explanation:
It is given that,
Length of the copper wire, l = 7300 m
Resistance of copper line, R = 10 ohms
Magnetic field, B = 0.1 T
Resistivity, 
We need to find the current flowing the copper wire. Firstly, we need to find the radius of he power line using physical dimensions as :
r = 0.00199 m
or
The magnetic field on a current carrying wire is given by :
I = 1000 A
So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.
The answer would be C. It will decrease with descent. Hope this helps!
Answer:
the ratio is 
Explanation:
Given
The RMS velocity of molecules in a gas is given by
where T=temperature
For T = 387K
For T = 774
dividing eqn 1 and eqn 2
Thus,the ratio is
Answer:
5.22 x 10^5 V
Explanation:
guessed on castle learning and got it right
Answer:
a) I = 13.04 A
b) R = 8.82 ohms
c) 1291.87 kilocalories are generated an hour.
Explanation:
let P be the power of the heater, V be the voltage of the heater, I be the current of the heater, R be the resistance.
a) we know that:
P = I×V
I = P/V
= (1500)/(115)
= 13.04 A
Therefore, the current of the heater is 13.04 A
b) we now have voltage and current, according to Ohm's law:
R = V/I
= (115)/(13.04)
= 8.82 ohms
Therefore, the resistance of the heating coil is 8.82 ohms.
c) the number of kilocalories generated in one hour by the heater is just the energy the heater produces in one hour which is given by:
E = P×t
= (1500)(1×60×60)
= 5400000 J
since 1 calorie = 4.81 J
1 kilocalorie = 0.001 calories
E = 5400000/4.18 ≈ 1291866.029 calories ≈1291.87 kilocalories
Therefore, 1291.87 kilocalories are produced/generated in one hour.
