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Elenna [48]
3 years ago
15

A circular loop of wire with 10 turns, radius 0.241 m and resistance 0.235 Ohms is connected to a 13.1 V power supply. The magne

tic field at the center of this loop is _____ T.
Physics
1 answer:
SVETLANKA909090 [29]3 years ago
8 0

Given Information:  

Resistance of circular loop = R = 0.235 Ω 

Radius of circular loop = r = 0.241 m

Number of turns = n = 10

Voltage = V = 13.1 V

Required Information:  

Magnetic field = B = ?  

Answer:  

Magnetic field = 0.00145 T

Explanation:  

In a circular loop of wire with n number of turns and radius r and carrying a current I induces a magnetic field B

B = μ₀nI/2r

Where μ₀= 4πx10⁻⁷ is the permeability of free space  and current in the loop is given by

I = V/R

I = 13.1/0.235

I = 55.74 A

B = 4πx10⁻⁷*10*55.74/2*0.241

B = 0.00145 T

Therefore, the magnetic field at the center of this circular loop is 0.00145 T

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nignag [31]

Answer:

speed of star is  1.37 × 10^{5} m/s

it is approaching to earth because wavelength of star is decreasing than rest

if emit same wavelength it does not move anywhere

it will remain steady condition with respect earth

Explanation:

given data

wavelength = 656.5 nm

rest wavelength = 656.8 nm

to find out

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solution

we know here equation that is

wavelength shift / wavelength at rest  = velocity of source / speed of light

so put all value and find

velocity of source = 3 × 10^{-9} × 3 × 10^{8} / 656.8 × 10^{-9}

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and

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3 years ago
Light travels at 3 × 108 m/s, and it takes about 8 min for light from the sun to travel to Earth. Based on this, the order of ma
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Answer:

The order of magnitude of the distance from the sun to Earth is 10⁸ km.

Explanation:

The order of magnitude of the distance from the sun to Earth can be calculated as follows:

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c: is the speed of light = 3x10⁸ m/s

t: is the time = 8 min

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I hope it helps you!

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3 years ago
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a rocket, initially at rest, is fired vertically with a net upward acceleration of 12 m/s2 . at an altitude of 0.50 km, the engi
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The rocket travelled a maximum height at 1.0102 km.

Given,

The acceleration of a rocket (a) = 12 m/s²

The altitude of the rocket (s) =  0.50 km = 0.5×10³m

The maximum height of the rocket (h) = ?

Solution,

A rocket is a spacecraft, aircraft, vehicle or projectile that obtains thrust from a rocket engine.

The rate of change of the velocity of an object with respect to time is known as acceleration. It is denoted by (a).i.e. unit is m/s²

(a) = Δv/Δt

Where , Δv is change in velocity and Δt is change in time.

The rate of change in position with respect to time is known as velocity. i.e. Its unit is m/s.

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Where,Δx is the change in position and Δt is change in time & v is velocity.

Therefore we know the equation of motion is written as,

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Then putting the value ,

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Therefore, at altitude of 0.50 km the initial velocity of rocket (u) will be 100 m/s, final velocity v become zero and under free falling the acceleration will be taken (-g) then equation of motion can be given as ,

v² = u² - 2(g)h

h = (v²- u² ) / 2g

h = 10,000/2×9.8

h = 510.2 m

So that the rocket travelled the maximum height ,

(h)= (0.5 km + 510.2m)

(h) = 1.0102 km

Hence, the rocket travelled at the maximum height h is 1.0102 km

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