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MAVERICK [17]
3 years ago
8

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
Physics
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

#LearnwithBrainly

11Alexandr11 [23.1K]3 years ago
6 0

Answer:

a  and  d

Explanation:

You might be interested in
A charge of 32.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of 4.30x 104 V/
hodyreva [135]

A) The work done by the electric field is zero

B) The work done by the electric field is 9.1\cdot 10^{-4} J

C) The work done by the electric field is -2.4\cdot 10^{-3} J

Explanation:

A)

The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).

The work done by a force is given by the equation

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement of the particle

\theta is the angle between the direction of the force and the direction of the displacement

In this problem, we have:

  • The force is directed vertically upward (because the field is directed vertically upward)
  • The charge moves to the right, so its displacement is to the right

This means that force and displacement are perpendicular to each other, so

\theta=90^{\circ}

and cos 90^{\circ}=0: therefore, the work done on the charge by the electric field is zero.

B)

In this case, the charge move upward (same direction as the electric field), so

\theta=0^{\circ}

and

cos 0^{\circ}=1

Therefore, the work done by the electric force is

W=Fd

and we have:

F=qE is the magnitude of the electric force. Since

E=4.30\cdot 10^4 V/m is the magnitude of the electric field

q=32.0 nC = 32.0\cdot 10^{-9}C is the charge

The electric force is

F=(32.0\cdot 10^{-9})(4.30\cdot 10^4)=1.38\cdot 10^{-3} N

The displacement of the particle is

d = 0.660 m

Therefore, the work done is

W=Fd=(1.38\cdot 10^{-3})(0.660)=9.1\cdot 10^{-4} J

C)

In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is

\theta=90^{\circ}+45^{\circ}=135^{\circ}

Moreover, we have:

F=1.38\cdot 10^{-3} N (electric force calculated in part b)

While the displacement of the charge is

d = 2.50 m

Therefore, we can now calculate the work done by the electric force:

W=Fdcos \theta = (1.38\cdot 10^{-3})(2.50)(cos 135.0^{\circ})=-2.4\cdot 10^{-3} J

And the work is negative because the electric force is opposite direction to the displacement of the charge.

Learn more about work and electric force:

brainly.com/question/6763771

brainly.com/question/6443626

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0
3 years ago
An elevator cable breaks when a 925-kg elevator is 28.5 m above the top of a huge spring Ak = 8.00 * 104 N????mB at the bottom o
rosijanka [135]

Answer:

a) = 258352.5J

b) = 23.63 m/s

c) = 1.8m

Explanation:

Data;

Mass = 925kg

Distance (s) = 28.5m

Force constant (k) = 8.0*10⁴ N/m

g = 9.8 m/s²

a) = work = force * distance

But force = mass * acceleration

Force = 925 * 9.8 = 9065N

Work = F * s = 9065 * 28.5 = 258352.5J

b) acceleration (a) = (v² - u²) / 2s

a = v² / 2s

v² = a * 2s

v² = 9.8 * (2 * 28.5)

v² = 9.8 * 57

v² = 558.6

v = √(558.6)

V = 23.63 m/s

C). The work stops when the work done to raise the spring equals the work done to stop it by the spring

W = ½kx²

258352.5 = ½ * 8.0*10⁴ * x²

(2 * 258352.5) = 8.0*10⁴x²

516705 = 8.0*10⁴x²

X² = 516705 / 8.0*10⁴

X² = 6.46

X = √(6.46)

X = 2.54m

The compression was about 2.54m

3 0
3 years ago
Read 2 more answers
A photon in a laboratory experiment has an energy of 5 eV. What is the frequency of this photon? (using the idea of the electron
andreyandreev [35.5K]
Answer and working shown on photo

7 0
3 years ago
When an automobile moves with constant speed down a highway, most of the power developed by the engine is used to compensate for
uysha [10]

Answer:

F=5449 N

Explanation:

Work done is a product of force and displacement ie

Work done, W, = Force*Displacement

Power, P, is Work done/Time

P=\frac {W}{t}=\frac {FS}{t} where P is power, W is work done, F is force, S is displacement and t is time

In this case, F is the frictional force. Converting the power from hp to W, we multiply by 746 hence P=746*168=125328  W

Since displacement/time is velocity, then

P=FV where V is velocity in m/s

Making F the subject

F=\frac {P}{V}

F=\frac {125328}{23}=5449.043478  N

F=5449 N

7 0
3 years ago
A 250 kg cart is at the top of a hill that is 32 m high, what is its potential energy?
atroni [7]

Answer:

<h2>80,000 J</h2>

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 10 m/s²

From the question we have

PE = 250 × 10 × 32

We have the final answer as

<h3>80,000 J</h3>

Hope this helps you

3 0
3 years ago
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