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3 years ago

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Which formulas have been correctly rearranged to solve for radius? Check all that apply. r = GM central/v^2 r =fcm/v^2 r =ac/v^2

r =vt/2pi r =act/pi

2 answers:

jek_recluse [69]3 years ago

7 0

The orbital radius is: $r=\frac{GM}{v^2}$

Explanation:

The problem is asking to find the radius of the orbit of a satellite around a planet, given the orbital speed of the satellite.

For a satellite in orbit around a planet, the gravitational force provides the required centripetal force to keep it in circular motion, therefore we can write:

$\frac{GMm}{r^2}=m\frac{v^2}{r}$

where

G is the gravitational constant

M is the mass of the planet

m is the mass of the satellite

r is the radius of the orbit

v is the speed of the satellite

Re-arranging the equation, we find:

$\frac{GM}{r}=v^2\\r=\frac{GM}{v^2}$

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

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11Alexandr11 [23.1K]3 years ago

6 0

Answer:

a and d

Explanation:

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A light ray incident from medium 1 to medium 2, where n1>n2. When the incident angle exceed the critical angle ac, the refrac

vovikov84 [41]

Explanation:

(a)

Critical angle is the angle at the angle of refraction is 90°. After the critical angle, no refraction takes place.

Using Snell's law as:

$n_1\times {sin\theta_i}={n_2}\times{sin\theta_r}$

Where,

${\theta_i}$ is the angle of incidence

${\theta_r}$ is the angle of refraction = 90°

${n_2}$ is the refractive index of the refraction medium

${n_1}$ is the refractive index of the incidence medium

Thus,

$n_1\times {sin\ \theta_{critical}}={n_2}\times{sin\ 90^0}$

The formula for the calculation of critical angle is:

${sin\theta_{critical}}=\frac {n_2}{n_1}$

Where,

${\theta_{critical}}$ is the critical angle

(b)

No it cannot occur. It only occur when the light ray bends away from the normal which means that when it travels from denser to rarer medium.

7 0

3 years ago

An aerobatic airplane pilot experiences weightlessness as she passes over the top of a loop-the-loop maneuver. The acceleration

Debora [2.8K]

Answer:

The radius of the loop is 20.66 km

Explanation:

let the radius of the loop be r

mass of airplane is m

At the top, the pilot experiences two radial forces, which are

1) Gravitational force is mg

2) Centrifugal forces mv²/r out of the center

When the pilot experiences no weight,

then, mg = mv²/r

r = v² / g

= 450² / 9.8

= 20.66 x 10³3

= 20.66 km

3 0

3 years ago

A ball traveling at 15 m/s hits a bat with a force of 200N. How much force does the bat (moving at 20m/s)

just olya [345]

Answer:

200 N

Explanation:

Given that,

A ball traveling at 15 m/s hits a bat with a force of 200 N.

We need to find the force that the bat moving at 20 m/s hit the ball with.

We know that, this probelm is based on Newton's third law of motion. The force that the ball exerting on bat should be equal to the force that the bat exerting in the ball but in opposite direction.

It would mean that the ball hits the ball with a force of 200 N. Hence, the correct option is (a).

8 0

3 years ago

A 1,000 kg car has 50,000 joules of kinetic energy what is its speed??

Vaselesa [24]

Using KE=1/2mv^2

v= $\sqrt{50000/500$

so v=10

3 0

3 years ago

Read 2 more answers

Andrea and Chuck are riding on a merry-go-round. Andrea rides on a horse at the outer rim of the circular platform, twice as far

marta [7]

a) Their angular speeds are the same

b) Andrea's tangential speed is twice the value of Chuck's tangential speed

Explanation:

a)

The angular speed of Andrea and Chuck is the same.

Let's call $\omega$ the angular speed at which the merry-go-round is rotating. We know that the angular speed is defined as:

$\omega= \frac{2\pi}{T}$

where

$2 \pi$ is the angular displacement covered in one revolution

T is the period of revolution

The merry go round is a rigid body, so all its point cover the same angular displacement in the same time: this means that it doesn't matter where Andrea and Chuck are located along the merry-go-round, their angular speed will still be the same.

b)

For an object in circular motion, the tangential speed is given by

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance from the centre of rotation

Here let's call $r_c$ the distance at which Chuck is rotating, so his tangential speed is

$v_c = \omega r_c$

Now we know that Andrea is rotating twice as far from the centre, so at a distance of

$r_a = 2 r_c$

So his tangential speed is

$v_a = \omega r_a = \omega (2 r_c) = 2(\omega r_c) = 2 v_c$

So, Andrea's tangential speed is twice the value of Chuck's tangential speed.

Learn more about circular motion:

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5 0

3 years ago