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3 years ago

Besides gravity, what factor keeps the moon and Earth in orbit?

Physics

1 answer:

Gelneren [198K]3 years ago

7 0

Answer:

(D) Inertia

Explanation:

Inertia words with Gravity to keep the Moon, Earth and Sun All in Orbit!

<em>-Aslina</em>

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Review. An electron moves in a circular path perpendicular to a constant magnetic field of magnitude 1.00 mT . The angular momen

Ghella [55]

The speed of an electron when it moves in a circular path perpendicular to a constant magnetic field is 8.88 x 10^7 m/s.

The angular momentum(L) of an electron moving in a circular path is given by the formula,

L = mvr ........(i)

We know that the radius of the path of an electron in a magnetic field is

r = mv/qB

Putting this value in equation (i),

L = mv x mv/qB

or L = (mv)^2/qB

Putting the given values in the above equation,

4 x 10^-25 = (9.1x10^-31)^2 x v^2/ 1.6 x 10^-19 x 1 x 10^-3

v comes out to be 8.88 x 10^7 m/s.

Hence, the speed of an electron when it moves in a circular path perpendicular to a constant magnetic field is 8.88 x 10^7 m/s.

To know more about "angular momentum", refer to the following link:

brainly.com/question/15104254?referrer=searchResults

#SPJ4

5 0

2 years ago

A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the

Margaret [11]

Hi there!

$\boxed{\omega = 0.38 rad/sec}$

We can use the conservation of angular momentum to solve.

$\large\boxed{L_i = L_f}$

Recall the equation for angular momentum:

$L = I\omega$

We can begin by writing out the scenario as a conservation of angular momentum:

$I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)$

$I_m$ = moment of inertia of the merry-go-round (kgm²)

$\omega_m$ = angular velocity of merry go round (rad/sec)

$\omega_f$ = final angular velocity of COMBINED objects (rad/sec)

$I_b$ = moment of inertia of boy (kgm²)

$\omega_b$ = angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

$I = MR^2$

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

$\omega = \frac{v}{r}$

$L_b = MR^2(\frac{v}{R}) = MRv$

Plug in the given values:

$L_b = (20)(3)(5) = 300 kgm^2/s$

Now, we must solve for the boy's moment of inertia:

$I = MR^2\\I = 20(3^2) = 180 kgm^2$

Use the above equation for conservation of momentum:

$600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}$

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3 years ago

Motion to the right is given a _________ sign, and motion to the left is given a negative __________.

siniylev [52]

Answer: Motion to the right is given a negative_________ sign, and motion to the left is given a negative velocity function __________.

Explanation:

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3 years ago

9. A van travels at a speed of 40.0 m/s has a mass of 600kg. What is the van’s kinetic energy? *

Zina [86]

$ke = \frac{1}{2} m{v}^{2} \\ = \frac{1}{2} (600)( {40}^{2} ) \\ = 300 \times 800 = 240000 = 240kjoule$

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3 years ago

How do we use the electromagnetic spectrum to determine how far things are in the universe and how they move?

BartSMP [9]

<span>The electromagnetic spectrum describes all the wavelengths of light. From dark nebulae to exploding stars, it reveals an otherwise invisible universe.</span>

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3 years ago