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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Given:
A cylindrical container closed of both end has a radius of 7cm and height of 6cm.
Explanation:
A.) Find the total surface area of the container.
- A = 2πrh + 2πr²
- A = 2(3.14)(7)(6) + 2(3.14)(7 × 7)
- A = 263.76 + 307.72
- A = 571.48
B.) Find the volume of the container.
- V = πr²h
- V = (3.14)(7×7)(6)
- V = 923.16
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Answer:
Explanation:
The relationship between the linear distance covered by an object and its angular displacement is given by the following formula:
s = rθ
where,
s = distance traveled on road = ?
r radius of tires = diameter/2 = 2.2 m/2 = 1.1 m
θ = angular displacement = (60 rev)(2π rad/1 rev) = 377 rad
Therefore,
The object's final velocity, given the data is 10.5 rad/s
<h3>What is acceleration? </h3>
This is defined as the rate of change of velocity which time. It is expressed as
a = (v – u) / t
Where
- a is the acceleration
- v is the final velocity
- u is the initial velocity
- t is the time
<h3>How to determine the final velocity</h3>
The following data were obtained from the question
- Initial velocity (u) = 1.5 rad/s
- Acceleration (a) = 0.75 rad/s²
- Time (t) = 12 s
- Final velocity (v) = ?
The final velocity can be obtained as follow:
a = (v – u) / t
0.75 = (v – 1.5) / 12
Cross multiply
v – 1.5 = 0.75 × 12
v – 1.5 = 9
Collect like terms
v = 9 + 1.5
v = 10.5 rad/s
Thus, the final velocity of the object is 10.5 rad/s
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Answer:
a
b
Explanation:
From the question we are told that
The pressure of the water in the pipe is 
The speed of the water is 
The original area of the pipe is 
The new area of the pipe is ![A_2 = \pi * \frac{[\frac{d}{2} ]^2}{4} = \pi * \frac{\frac{d^2}{4} }{4} = \pi \frac{d^2}{16}](https://tex.z-dn.net/?f=A_2%20%3D%20%5Cpi%20%2A%20%20%5Cfrac%7B%5B%5Cfrac%7Bd%7D%7B2%7D%20%5D%5E2%7D%7B4%7D%20%20%3D%20%20%5Cpi%20%2A%20%20%5Cfrac%7B%5Cfrac%7Bd%5E2%7D%7B4%7D%20%7D%7B4%7D%20%3D%20%5Cpi%20%5Cfrac%7Bd%5E2%7D%7B16%7D)
Generally the continuity equation is mathematically represented as
Here 
is the new velocity
So
=> 
=> 
=> 
=>
Generally given that the height of the original pipe and the narrower pipe are the same , then we will b making use of the Bernoulli's equation for constant height to calculate the pressure
This is mathematically represented as
Here 
is the density of water with value 
![P_2 = P_1 + \frac{1}{2} * \rho [ v_1^2 - v_2^2 ]](https://tex.z-dn.net/?f=P_2%20%3D%20%20P_1%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%2A%20%20%5Crho%20%5B%20v_1%5E2%20-%20v_2%5E2%20%5D)
=> ![P_2 = 110 *10^{3} + \frac{1}{2} * 1000 * [ 1.4 ^2 - 5.6 ^2 ]](https://tex.z-dn.net/?f=P_2%20%3D%20%20110%20%2A10%5E%7B3%7D%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%2A%20%201000%20%2A%20%20%5B%201.4%20%5E2%20-%205.6%20%5E2%20%5D)
=>
