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Marina CMI [18]
3 years ago
6

The following diagrams show five pairs of asteroids, labeled with their relative masses (M) and distances (d) between them. For

example, an asteroid with M=2 has twice the mass of one with M=1 and a distance of d=2 is twice as large as a distance of d=1. Rank each pair from left to right based on the strength of the gravitational force attracting the asteroids to each other, from strongest to weakest.
Physics
1 answer:
serg [7]3 years ago
7 0

Answer:

- d = 2; m = 2; m = 2

- d= 1; m = 1; m = 2

- d = 1; m = 1; m = 1

- d = 2; m = 1; m = 2

- d = 2; m = 1; m = 1

Explanation:

The following diagrams show five pairs of asteroids, labeled with their relative masses (M) and distances (d) between them. For example, an asteroid with M=2 has twice the mass of one with M=1 and a distance of d=2 is twice as large as a distance of d=1. Rank each pair from left to right based on the strength of the gravitational force attracting the asteroids to each other, from strongest to weakest.

- d = 2; m = 2; m = 2

- d= 1; m = 1; m = 2

- d = 1; m = 1; m = 1

- d = 2; m = 1; m = 2

- d = 2; m = 1; m = 1

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ale4655 [162]

Answer:

a.) The electric and magnetic fields are in phase with each other as they propagate through space.

Explanation:

Electromagnetic wave is a transverse wave in which magnetic field and electric field both induces each other as both changes with time

Here magnetic field induces electric field and similarly magnetic field induces electric field.

As we know that this is a transverse wave so here magnetic field and electric field lies in perpendicular planes. but they both propagate in same direction in such a wave that both fields reaches their maximum position and minimum positions simultaneously

So the correct answer is

a.) The electric and magnetic fields are in phase with each other as they propagate through space.

6 0
3 years ago
A 0.07 kg tennis ball, initially at rest, leaves a racket with a speed of 56 m/s. If the ball is in contact with
Naddika [18.5K]

The average force on the ball by the racket is 98 N. The correct option is the third option - 98 N

From the question, we are to determine the average force on the ball by the racket.

From the formula,

F = \frac{mv}{t}

Where F is the force

m is the mass

v is the velocity

and t is the time

From the given information

m = 0.07 kg

v = 56 m/s

t = 0.04 s

Putting the parameters into the formula,

we get

F = \frac{0.07 \times 56}{0.04}

F = \frac{3.92}{0.04}

F = 98 N

Hence, the average force on the ball by the racket is 98 N. The correct option is the third option - 98 N

Learn more on calculating force exerted on an object here: brainly.com/question/13590154

4 0
2 years ago
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Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant
irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

a_g = \frac{GM}{R^2}

Here

M = \text{Mass inside the Orbit of the star}

R = \text{Orbital radius}

G = \text{Universal Gravitational Constant}

Mass inside the orbit in terms of Volume and Density is

M =V \rho

Where,

V = Volume

\rho =Density

Now considering the volume of the star as a Sphere we have

V = \frac{4}{3} \pi R^3

Replacing at the previous equation we have,

M = (\frac{4}{3}\pi R^3)\rho

Now replacing the mass at the gravitational acceleration formula we have that

a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho

a_g = \frac{4}{3} G\pi R\rho

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration.  So centripetal acceleration of the star is

a_c = \frac{4}{3} G\pi R\rho

At the same time the general expression for the centripetal acceleration is

a_c = \frac{\Theta^2}{R}

Where \Theta is the orbital velocity

Using this expression in the left hand side of the equation we have that

\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2

\Theta = (\frac{4}{3}G\pi \rho R^2)^{1/2}

\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R

Considering the constant values we have that

\Theta = \text{Constant} \times R

\Theta \propto R

As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

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elena55 [62]

Answer:

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Explanation:

HOPE IT HELPS!!

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Answer:

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Explanation:

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