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3 years ago

Question 7 of 10

Physics

1 answer:

jok3333 [9.3K]3 years ago

4 0

Answer:

when a magnet is hanged freely in air it turns in the direction of the north and south while the magnetic north pole faces the south pole of the earth and magnetic south pole faces the north pole if the earth

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What is the definition for relative motion

xeze [42]

Relative motion is the calculation of the motion of an object with regard to some other moving object.

7 0

3 years ago

An exoplanet is in an elliptical orbit around a distant star. At its closest approach, the exoplanet is 0.540 AU from the star a

atroni [7]

Answer:

0.71121 km/s

Explanation:

$v_1$ = Velocity of planet initially = 54 km/s

$r_1$ = Distance from star = 0.54 AU

$v_2$ = Final velocity of planet

$r_2$ = Final distance from star = 41 AU

As the angular momentum of the system is conserved

$mv_1r_1=mv_2r_2\\\Rightarrow v_1r_1=v_2r_2\\\Rightarrow v_2=\frac{v_1r_1}{r_2}\\\Rightarrow v_2=\frac{54\times 0.54}{41}\\\Rightarrow v_2=0.71121\ km/s$

When the exoplanet is at its farthest distance from the star the speed is 0.71121 km/s.

7 0

3 years ago

Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J =

Sloan [31]

Answer:

The current is $I = 8.9 *10^{-5} \ A$

Explanation:

From the question we are told that

The radius is $r = 3.17 \ mm = 3.17 *10^{-3} \ m$

The current density is $J = c\cdot r^2 = 9.00*10^{6} \ A/m^4 \cdot r^2$

The distance we are considering is $r = 0.5 R = 0.001585$

Generally current density is mathematically represented as

$J = \frac{I}{A }$

Where A is the cross-sectional area represented as

$A = \pi r^2$

=> $J = \frac{I}{\pi r^2 }$

=> $I = J * (\pi r^2 )$

Now the change in current per unit length is mathematically evaluated as

$dI = 2 J * \pi r dr$

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows

$I = 2\pi \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr$