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.
Learn more about rms current here:-brainly.com/question/20913680
#SPJ4
Answer:
a) The current is i = 1.2 A
b) The charge is Q = 17280 C
c) The energy is E = 43200 J
Explanation:
a) The current is given by the ohm's law wich is:
i = V/R = 3/2.5 = 1.2 A
b) Since the charge is steady we can use the following equation to find the charge amount in that time:
i = Q/t
Q = t*i
Where t is in seconds, so we have 4h * 3600 = 14400 s
Q = 1.2*14400 = 17280 C
c) The energy is the power delivered to the toy multiplied by the time:
P = 1.2*2.5 = 3 W
E = P*t = 3*14400 = 43200 J
*may be egg shaped
*has no new stars being formed
*has almost no gas or dust between stars
I literally just took notes on elliptical galaxies a week ago wow
Answer:
Given that
V2/V1= 0.25
And we know that in adiabatic process
TV^န-1= constant
So
T1/T2=( V1 /V2)^ န-1
So = ( 1/0.25)^ 0.66= 2.5
Also PV^န= constant
So P1/P2= (V2/V1)^န
= (1/0.25)^1.66 = 9.98
A. RMS speed is
Vrms= √ 3RT/M
But this is also
Vrms 2/Vrms1= (√T2/T1)
Vrms2=√2.5= 1.6vrms1
B.
Lambda=V/4π√2πr²N
So
Lambda 2/lambda 1= V2/V1 = 0.25
So the mean free path can be inferred to be 0.25 times the first mean free path
C. Using
Eth= 3/2KT
So Eth2/Eth1= T2/T1
So
Eth2= 2.5Eth1
D.
Using CV= 3/2R
Cvf= Cvi
So molar specific heat constant does not change
Answer:
Technician A is right. The situation will happens even with only two bulbs in series
Explanation:
We must take into account that
1.- All electric device need its nominal voltage to operate
2.-Any and all electric device means an electric load for the source in terms of equation that means any device will implies a drop voltage of V = I*R ( I the flows current and R the resistance of the device)
3.-Nominal voltage for bulbs are specify for houses voltages you find between fase and neutral wires for instance in Venezuela 120 (v).
4.-In a imaginary circuit of only one bulb, the nominal voltage will be applied and the bulb will operates correctly, but when you add another bulb (in series) the nominal voltage will split between the two bulbs ( we could find a situation such as the first bulb work properly but the second one does not). The voltage split according to Ohms law (in such way that the sum of voltage between the terminal of the first bulb plus the voltage at terminals of the second one are equal to nominal voltage.
For that reason all the bulbs are connected in parallel in wich case all of them will operate with the common voltage