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stealth61 [152]
2 years ago
15

Many industries are powered via distant power stations. Calculate the current flowing through a 7,300m long 10. copper power lin

e that produces a 0.10T magnetic field. Note: No = 4 TX10-7T-m/A and copper = 1.72 x 10-82.m. Note: Do not assume the voltage transmitted on the power line. Note: This is a two-step problem: This is a two-step problem: Calculate the radius of the power line from the physical dimensions and then calculate the current from the magnetic field. A. 125A OB. 250A C.500A OD. 750A E. 1,000A
Physics
1 answer:
Oliga [24]2 years ago
4 0

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

\mu_o=4\pi \times 10^{-7}\ T-m/A

Resistivity, \rho=1.72\times 10^{-8}\ \Omega-m

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=\rho \dfrac{l}{A}

R=\rho \dfrac{l}{\pi r^2}

r=\sqrt{\dfrac{\rho l}{R\pi}}

r=\sqrt{\dfrac{1.72\times 10^{-8}\times 7300}{10\pi}}

r = 0.00199 m

or

r=1.99\times 10^{-3}\ m=2\times 10^{-3}\ m

The magnetic field on a current carrying wire is given by :

B=\dfrac{\mu_o I}{2\pi r}

I=\dfrac{2\pi rB}{\mu_o}

I=\dfrac{2\pi \times 0.1\times 2\times 10^{-3}}{4\pi \times 10^{-7}}

I = 1000 A

So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.

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A cannonball is fired perfectly horizontally from the top of a 210 m tall cliff. It is fired with an initial velocity of 50 m/s.
pochemuha

Answer:

the horizontal distance covered by the cannonball before it hits the ground is 327.5 m

Explanation:

Given;

height of the cliff, h = 210 m

initial horizontal velocity of the cannonball, Ux = 50 m/s

initial vertical velocity of the cannonball, Uy = 0

The time for the cannonball to reach the ground is calculated as;

h = u_yt - \frac{1}{2} gt^2\\\\h = 0 - \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 210}{9.8} }\\\\t  = 6.55 \ s

The horizontal distance covered by the cannonball before it hits the ground is calculated as;

X = U_x \times \ t\\\\X = 50 \times \ 6.55\\\\X = 327.5 \ m

Therefore, the horizontal distance covered by the cannonball before it hits the ground is 327.5 m

8 0
2 years ago
Identify the forces acting on the cars and explain why the cars do not slide down the hill
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5 0
3 years ago
Use what you know about reflection and absorption of light to complete the sentence.
Blababa [14]

A red ladybug appears red in white light, red in red light, and black in blue light. Those would be the proper selections you'd need.

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3 years ago
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. Since the man is walking at a constant velocity, he has what acceleration
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Two astronauts, each having a mass of 74.3 kg are connected by a 13.1 m rope of negligible mass. They are isolated in space, orb
murzikaleks [220]

Answer:

  L = 5076.5 kg m² / s

Explanation:

The angular momentum of a particle is given by

         L = r xp

         L = r m v sin θ

the bold are vectors, where the angle is between the position vector and the velocity, in this case it is 90º therefore the sine is 1

as we have two bodies

       L = 2 r m v

let's find the distance from the center of mass, let's place a reference frame on one of the masses

        x_{cm} = \frac{1}{M} \sum  x_{i} m_{i}i

        x_{cm} = \frac{1}{m+m} ( 0 + l m)

        x_{cm} = \frac{1}{2m}  lm

        x_{cm} = \frac{1}{2}

        x_{cm} = 13.1 / 2 = 6.05 m

let's calculate

          L = 2  6.05  74.3  5.65

          L = 5076.5 kg m² / s

4 0
3 years ago
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