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stealth61 [152]
3 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]3 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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Charge X has twice as much charge as particle Y. The two charges are placed near each other. Compared to the force on particle X
Art [367]

The force on charge Y is the same as the force on charge X

Explanation:

We can answer this problem by applying Newton's third law of motion, which states that:

"When an object A exerts a force on object B (action force), then object B exerts an equal and opposite force on object A (reaction force)"

In this problem, we can identify object A as charge X and object B as charge Y. The magnitude of the electrostatic force between them is given by

F=k\frac{q_x q_y}{r^2}  (1)

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_x, q_y are the two charges

r is the separation between the two charges

According to Newton's third law, therefore, the magnitude of the force exerted by charge X on charge Y is the  same as the force exerted by charge Y on charge X (and it is given by eq.(1)), however their directions are opposite.

Learn more about Newton's third law:

brainly.com/question/11411375

#LearnwithBrainly

6 0
3 years ago
Two point charges each carrying a charge of + 4.5 E - 6 C are located 4.5 meters away from each other. How strong is the electro
BARSIC [14]

Answer:

0.009 N, repulsive

Explanation:

The electrostatic force between two electric charges is given by:

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1 and q2 are the two charges

r is the separation between the two charges

In this problem, we have

q_1 =q_2 = +4.5\cdot 10^{-6}C are the two charges

r = 4.5 m is their separation

Substituting into the equation, we find

F=(9\cdot 10^9 Nm^2 C^{-2})\frac{(+4.5\cdot 10^{-6} C)(4.5\cdot 10^{-6} C)}{(4.5 m)^2}=0.009 N

Moreover, the force is repulsive. In fact, the following rules apply:

- When two charges have same sign, they repel each other

- When two charges have opposite signs, they attract each other

7 0
3 years ago
A truck traveled 400 meters north in 80 seconds and then it traveled 300 meters east in 70 seconds. The magnitude of the average
Varvara68 [4.7K]

Answer:

3.3m/s

Explanation:

You first get the total time (80 + 70 = 150s).

Then you would find the displacement of the truck. To do that you would do component method (vector addition), so since its a right triangle (North and East), displacement is 400^2 + 300^2 = d^2.

d= 500m.

So now that you have displacement and time, you can find the velocity:

v=d/t

v=500/150

v=3.3

5 0
3 years ago
Calculate (a) the torque, (b) the energy, and (c) the average power required to accelerate Earth in 4.0 days from rest to its pr
natima [27]
<h2>Answer:</h2>

Torque = <em>2.05 x 10²⁸ Nm</em>

Energy = <em>3.54 x 10³³ J</em>

Average power = <em>1.02 x 10²⁸ W</em>

<h2>Explanation:</h2>

(a) Torque (τ) is the rotational effect of a given force.  

It is given by

τ = I x α          -------------(i)

Where;

I = rotational inertia of the object

α = angular acceleration of the object.

In this case, the object is the Earth. Therefore,

I = 9.71 x 10³⁷ kg m²

α = ω / t

Where;

ω = angular velocity of earth = 2π rad / day

<em>Since </em>

<em>1 day = 24 hours and 1 hour = 3600seconds</em>

<em>1 day = 24 x 3600 seconds = 86400seconds</em>

<em>=> ω = 2π rad / 86400seconds</em>

<em>=> ω = 7.29 × 10⁻⁵ rad/s</em>

<em />

t = 4 days = 4 x 24 x 3600 seconds = 345600 seconds

=> α = ω / t

=> α = 7.29 × 10⁻⁵ / 345600

=> α = (7.29 × 10⁻⁵) / (3.456 x 10⁵)

=> α = (7.29 × 10⁻⁵⁻⁵) / (3.456)

=> α = (7.29 × 10⁻¹⁰) / (3.456)

=> α = 2.11 × 10⁻¹⁰ rad/s²

Now substitute the values of I and α into equation (i)

τ = 9.71 x 10³⁷ x 2.11 × 10⁻¹⁰

τ = 9.71 x 10²⁷ x 2.11

τ = 20.5 x 10²⁷ Nm

τ = 2.05 x 10²⁸ Nm

(ii) The energy (rotational energy) E is given by;

E = \frac{1}{2} x I x ω

E = \frac{1}{2} x 9.71 x 10³⁷ x 7.29 × 10⁻⁵

E = 35.4 x 10³² J

E = 3.54 x 10³³ J

(iii) The average power P, is given by;

P = E / t

P = 3.54 x 10³³ / 345600

P = 1.02 x 10²⁸ W

5 0
3 years ago
This time particle A starts from rest and accelerates to the right at 65.5 cm/s
FrozenT [24]

Answer:

t = 4 s

Explanation:

As we know that the particle A starts from Rest with constant acceleration

So the distance moved by the particle in given time "t"

d = v_i t + \frac{1}{2}at^2

d = 0 + \frac{1}{2}(65.5)t^2

d_1 = 32.75 t^2 cm

Now we know that B moves with constant speed so in the same time B will move to another distance

d_2 = 44 \times t

now we know that B is already 349 cm down the track

so if A and B will meet after time "t"

then in that case

d_1 = 349 + d_2

32.75 t^2 = 349 + 44 t

on solving above kinematics equation we have

t = 4 s

4 0
3 years ago
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