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Katen [24]
3 years ago
10

A high power line carries a current of 1.0 kA. What is the strength of the magnetic field this line produces at the ground, 10 m

away? ( μ 0 = 4π × 10 -7 T ∙ m/A)
Physics
1 answer:
solmaris [256]3 years ago
8 0

Answer:

The strength of the magnetic field that the line produces is 2x10^{-5} Tesla.

Explanation:

From Biot-Savart law, the equation to determine the strength of the magnetic field for any straight wire can be deduced:

           

B = \frac{\mu_{0}I}{2\pi r} (1)      

                                     

Where \mu_{0} is the permiability constant, I is the current and r is the distance from the wire.    

             

Notice that it is necessary to express the current, I, from kiloampere to ampere.

I = 1.0kA \cdot \frac{1000A}{1kA} ⇒ 1000A

Finally, equation 1 can be used:

B = \frac{(4\pi x10^{-7}T.m/A)(1000A)}{2\pi (10m)}    

           

B = 2x10^{-5}T    

Hence, the strength of the magnetic field that the line produces is 2x10^{-5} Tesla.

         

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Consider the space between a point charge and the surface of a neutral spherical conducting shell. If the charge sits at the cen
Furkat [3]

Answer:

True

Explanation:

If a thin, spherical, conducting shell carries a negative charge, We expect the excess electrons to mutually repel one another, and, thereby, become uniformly distributed over the surface of the shell. The electric field-lines produced outside such a charge distribution point towards the surface of the conductor, and end on the excess electrons. Moreover, the field-lines are normal to the surface of the conductor. This must be the case, otherwise the electric field would have a component parallel to the conducting surface. Since the excess electrons are free to move through the conductor, any parallel component of the field would cause a redistribution of the charges on the shell. This process will only cease when the parallel component has been reduced to zero over the whole surface of the shell

According to Gauss law

∅ = EA =-Q/∈₀

Where ∅  is the electric flux through the gaussian surface and E is the electric field strength

If the gaussian surface encloses no charge, since all of the charge lies on the shell, so it follows from Gauss' law, and symmetry, that the electric field inside the shell is zero. In fact, the electric field inside any closed hollow conductor is zero

8 0
3 years ago
Explain how energy is transferred in an impact situation such as a car crash.
DaniilM [7]
During a car crash, energy is transferred from the vehicle to whatever it hits, be it another vehicle or a stationary object. ... The object that was struck will either absorb the energy thrust upon it or possibly transfer that energy back to the vehicle that struck it.

I HOPE THIS HELPSS???
Mark me brainliest
8 0
3 years ago
Which is not true of the Intertropical Convergence Zone?A) It features heavy precipitation B) It's where the trade winds collide
olasank [31]
<h2>Answer: C) It's a high-pressure zone with sinking air</h2>

Explanation:

The intertropical convergence zone is the region of the terrestrial globe where the trade winds of the northern hemisphere converge with those of the southern hemisphere.  

It is characterized by being <u>a belt of low pressure</u> and inconsistent location around the equator constituted by ascending air currents, where large masses of warm and humid air converge from the north and south of the intertropical zone.  

The reason of its inconsistent location is due to the movements of the Earth with the seasons, having as a consequence the amount variation of heat energy from the sun in this region.

7 0
3 years ago
A car travelling at a constant speed of 70km/h passes a stationary police car. The police car immediately goes on the chase acce
Virty [35]

Answer:

18.24 seconds

Explanation:

First you convert the km/h to m/s, 70km/h=(175/9)m/s,85km/h=(425/18)m/s.

You know it took 10 seconds for the police to reach 85 km/h. Calculate the distance that the car is ahead of the police (175/9)*10=1750/9m. Then by divide 1750/9 with 425/18, you will get the value 8.24. Add the 10 seconds with the 8.24 you will get 18.24 sec which is the total time.

5 0
3 years ago
Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d
Tasya [4]

Answer:

I(x)  = 1444×k ×{\pi}

I(y)  = 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}  

Explanation:

Given data

function =  x^2 + y^2 ≤ 36

function =  x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to \pi /2 for θ

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} y²p dA

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} (a sinθ)²(k × a²) adθda

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (sin²θ)dθ

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (1-cos2θ)/2 dθ

I(x)  = k ({r}^{6}/6)^(5)_0 ×  {θ/2 - sin2θ/4}^{\pi /2}_0

I(x)  = k × ({6}^{6}/6) × (  {\pi /4} - sin\pi /4)

I(x)  = k ×  ({6}^{5}) ×   {\pi /4}

I(x)  = 1444×k ×{\pi}    .....................1

and we can say I(x) = I(y)   by the symmetry rule

and here I(o) will be  I(x) + I(y) i.e

I(o) = 2 × 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}   ......................2

3 0
3 years ago
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