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3 years ago

HURRY!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Physics

2 answers:

Paha777 [63]3 years ago

6 0

Light sound and thermal

drek231 [11]3 years ago

6 0

D.) light, sound, and thermal

Hope this helps!

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How do you answer number 3 a and b?

netineya [11]

I would assume you answer them by doing the math required.

7 0

4 years ago

A 200-loop coil of cross sectional area 8.5 cm2 lies in the plane of the paper. Directed out of the plane of the paper is a magn

Maslowich

Explanation:

It is given that,

Number of turns, N = 200

Area of cross section, A = 8.5 cm²

Magnetic field is directed out of the paper and is, B = 0.06 T

The magnetic field is out of the paper decreases to 0.02 T in 12 milliseconds. We need to find the direction of current induced. The induced emf is given by :

$\epsilon=-N\dfrac{d\phi}{dt}$

Since, $\epsilon=IR$

I is the induced current

$I=-\dfrac{N}{R}\dfrac{d\phi}{dt}$

According to Lenz's law, the direction of induced current is such that it always opposes the change in current that causes it.

Here, the field is directed out of the plane of the paper, this gives the induced current in counterclockwise direction.

5 0

3 years ago

Which is true of electricity generated both from coal and from nuclear reactions?

Fynjy0 [20]

Neither source is a renewable one.

6 0

3 years ago

If it has enough kinetic energy, a molecule at the surface of the Earth can "escape the Earth's gravitation", in the sense that

user100 [1]

Answer:

$K_E=mgr_E$

Explanation:

By conservation of energy, the sum of the kinetic and gravitational potential energies at the surface of the Earth must be equal than their sum at infinity, so we have:

$K_E+U_E=K_\infty+U_\infty$

$\frac{mv_E^2}{2}-\frac{GM_Em}{r_E}=\frac{mv_\infty^2}{2}-\frac{GM_Em}{r_\infty}$

Where $G=6.67\times10^{-11}Nm^2/kg^2$ is the gravitational constant, $M_E=5.97\times10^{24}kg$ and $r_E=6371000m$ are the mass and radius of the Earth, m is the mass of the particle, $v_E$ its velocity at the surface of the Earth (which would be its escape velocity) and $v_\infty$ and $r_\infty$ are the velocities and distance at infinity, which would be null and infinity respectively, so the right hand side of our equation is 0J, which leaves us with:

$\frac{GM_Em}{r_E}=\frac{mv_E^2}{2}=K_E$

Also, since the force the molecule experiments is the force of gravity (disregarding drag), we can write its weight in terms of Newton's Law of Gravitation:

$F=mg=\frac{GM_Em}{r_E^2}$