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KATRIN_1 [288]
3 years ago
6

Kepler's second law states that

Physics
1 answer:
mafiozo [28]3 years ago
7 0
Awnser A. The orbits of the planets are
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What will oppose most forces and slow objects
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4 years ago
A charge Q is distributed uniformly around the perimeter of a ring of radius R. Determine the electric potential difference betw
il63 [147K]

Answer:

the electric potential difference between the point at the center of the ring and a point on its axis ΔV is ( 0.8356 )\frac{kQ}{R}

Explanation:

Given the data in the question;

electric potential at the center of the ring V₀ = kQ / R

electric potential on the axis point Vr = kQ / √( R² + x² )

at a distance 6R from the center,

point at x = 6R

so distance circumference r = √( R² + (6R)² )

so

electric potential on the axis point Vr = kQ / √( R² + (6R)² )

Vr = kQ / R√37

Now

ΔV = V₀ - Vr

we substitute

ΔV = ( kQ / R) - ( kQ / R√37 )

ΔV =  kQ/R( 1 - 1/√37 )

ΔV =  kQ/R( 1 - 0.164398987 )

ΔV =  kQ/R( 0.8356 )

ΔV = ( 0.8356 )\frac{kQ}{R}       { where k = \frac{1}{4\pi e_0} }

Therefore, the electric potential difference between the point at the center of the ring and a point on its axis ΔV is ( 0.8356 )\frac{kQ}{R}

5 0
3 years ago
PLZ WHAT I DOO<br> I BROKE A WINDOW WHAT I DO
Hatshy [7]

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3 0
3 years ago
A 100 A current circulates around a 2.00-mm-diameter superconducting ring.
igomit [66]

Answer:

0.000314 Am²

6.049*10^-7 T

Explanation:

A

From the definitions of magnetic dipole moment, we can establish that

= , where

= the magnetic dipole moment in itself

= Current, 100 A

= Area, πr² (r = diameter divided by 2). Converting to m², we have 0.000001 m²

On solving, we have

= ,

= 100 * 3.14 * 0.000001

= 0.000314 Am²

B

= (0)/4 * 2 / ³, where

(0) = constant of permeability = 1.256*10^-6

z = 4.7 cm = 0.047 m

B = 1.256*10^-6 / 4*3.142 * [2 * 0.000314/0.047³]

B = 1*10^-7 * 0.000628/1.038*10^-4

B = 1*10^-7 * 6.049

B = 6.049*10^-7 T

8 0
3 years ago
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