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3 years ago

A place that limits a wave’s motion.

Physics

1 answer:

Yuki888 [10]3 years ago

7 0

Answer:

boundary

Explanation:

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What's true about the elliptical path that the planets follow around the sun? A. A line can be drawn from the planet to the sun

wariber [46]

Answer:

A. A line can be drawn from the planet to the sun that sweeps out equal areas in equal times

Explanation:

This is exactly what Kepler's second law of planetary motion states:

"the segment joining the sun with the center of each planet sweeps out equal areas in equal time"

This law basically tells how the speed of a planet orbiting the sun changes during its revolution. In fact, we have that:

- when a planet is closer to the Sun, it will orbit faster

- when a planet is farther from the Sun, it will orbit slower

5 0

4 years ago

By what factor is the self-inductance of an air solenoid changed if its length and number of coil turns are both tripled

fredd [130]

Answer:

The new self inductance is 3 times of the initial self inductance.

Explanation:

The self inductance of a solenoid is given by :

$L=\dfrac{\mu_oN^2 A}{L}$

Where

N is number of turns per unit length

A is area of cross section

l is length of solenoid

If length and number of coil turns are both tripled,

l' = 3l and N' = 3N

New self inductance is given by :

$L'=\dfrac{\mu_oN'^2 A}{L'}\\\\=\dfrac{\mu_o(3N)^2 A}{3L}\\\\=3\dfrac{\mu_oN^2 A}{L}\\\\=3L$

So, the new self inductance is 3 times of the initial self inductance.

4 0

3 years ago

You stand on a frictional platform that is rotating at 1.8 rev/s. Your arms are outstretched, and you hold a heavy weight in eac

dusya [7]

Answer:

20.62361 rad/s

489.81804 J

Explanation:

$I_i$ = Initial moment of inertia = 9.3 kgm²

$I_f$ = Final moment of inertia = 5.1 kgm²

$\omega_i$ = Initial angular speed = 1.8 rev/s

$\omega_f$ = Final angular speed

As the angular momentum of the system is conserved

$I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_f=\dfrac{I_i\omega_i}{I_f}\\\Rightarrow \omega_f=\dfrac{9.3\times 1.8}{5.1}\\\Rightarrow \omega_f=3.28235\ rev/s=3.28235\times 2\pi=20.62361\ rad/s$

The resulting angular speed of the platform is 20.62361 rad/s

Change in kinetic energy is given by

$\Delta K=\dfrac{1}{2}(I_f\omega_f^2-I_i\omega_i^2)\\\Rightarrow \Delta K=\dfrac{1}{2}(5.1\times (20.62361)^2-9.3\times (1.8\times 2\pi)^2)\\\Rightarrow \Delta K=489.81804\ J$