The rms current in the transmission lines is I = 487.18 A.
The root-imply-rectangular (rms) voltage of a sinusoidal supply of electromotive force is used to represent the source. it is the rectangular root of the time average of the voltage squared.
Alternating-present day circuits. the root-imply-square (rms) voltage of a sinusoidal source of electromotive force is used to symbolize the supply. it's far the square root of the time average of the voltage squared.
Electric power is by using present day or the waft of electric fee and voltage or the capacity of rate to deliver electricity. A given cost of power can be produced by using any combination of contemporary and voltage values
power = 38 M watt
rms voltage = 78 K v
power = IV
I = power/V
I = (38 * 1000000)/78*1000
I = 487.18 A.
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I think analog but I could be wrong
Answer:
Explanation:
Far point = 17 cm . That means he can not see beyond this distance .
He wants to see at an object at 65 cm away . That means object placed at 65 has image at 17 cm by concave lens . Using lens formula
1 / v - 1 / u = 1 / f
1 / - 17 - 1 / - 65 = 1 / f
= 1 / 65 - 1 / 17
= - .0434 = 1 / f
power = - 100 / f
= - 100 x .0434
= - 4.34 D .
This question is based on the fundamental assumption of vector direction.
A vector is a physical quantity which has magnitude as well direction for its complete specification.
The magnitude of a physical quantity is simply a numerical number .Hence it can not be negative.
A negative vector is a vector which comes into existence when it is opposite to our assumed direction with respect to any other vector. For instance, the vector is taken positive if it is along + X axis and negative if it is along - X axis.
As per the first option it is given that a vector is negative if its magnitude is greater than 1. It is not correct as magnitude play no role in it.
The second option tells that the magnitude of the vector is less than 1. Magnitude can not be negative. So this is also wrong.
Third one tells that a vector is negative if its displacement is along north. It does not give any detail information about the negativity of a vector.
In a general sense we assume that vertically downward motion is negative and vertically upward is positive. In case of a falling object the motion is vertically downward. So the velocity of that object is negative .
So last option is partially correct as the vector can be negative depending on our choice of co-ordinate system.
Answer:
Please find the complete solution in the attached file.
Explanation:
