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2 years ago

Which variables are involved in understanding Kepler's third law of motion? (1 point)

Physics

2 answers:

Zielflug [23.3K]2 years ago

7 0

The variables which are involved in understanding Kepler's third law of

motion are

Orbital velocity

Distance to sun

<h3 /><h3>What is Kepler's third law of motion?</h3>

Kepler's third law of motion states that the the square of the orbital period of

a planet is proportional to the cube of the semi-major axis of its orbit. He

also inferred that the greater the distance, the slower the orbital velocity.

This thereby makes option D the most appropriate option as it contains the

orbital velocity and distance to sun variables.

Read more about Kepler's third law of motion here brainly.com/question/777046

klasskru [66]2 years ago

5 0

Answer:orbital velocity distance to sun

Explanation:

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What is the net work done on the 20kg block while it moves the 4 meters?

Vinil7 [7]

Answer:

The answer to your question is 784.8 J. None of your answer, did you forget some information?

Explanation:

Data

mass = 20 kg

distance = 4 m

work = ?

Formula

Work = force x distance

Force = mass x gravity

Process

1.- Calculate the weight of the block

Weight = 20 x 9.81

Weight = 196.2 N

2.- Calculate the work done

Work = 196.2 x 4

Work = 784.8 J

5 0

3 years ago

Professional Application. A 96 kg football player catches a 0.900 kg ball with his feet off the ground with both of them moving

Zarrin [17]

To solve this problem it is necessary to apply the equations related to the conservation of momentum.

This definition can be expressed as

$m_1u_1+m_2u_2 = (m_1+m_2)V_f$

Where

$m_{1,2}$ = Mass of each object

$u_{1,2}$ = Initial Velocity of each object

$V_f$ = Final velocity

Rearranging the equation to find the final velocity we have,

$V_f = \frac{m_1u_1+m_2u_2}{(m_1+m_2)}$

Our values are given as

$m_1 = 96Kg\\m_2 = 0.9Kg\\u_1 = 6.3m/s\\u_2 = 27.4m/s$

Replacing we have,

$V_f = \frac{(96)(6.3)+(0.9)(27.4)}{(96+0.9)}$

$V_f = 6.4959m/s$

Therefore the final velocity is 6.5m/s

3 0

3 years ago

When a 2.50 kg object is hung vertically on a certain light spring described by hookes law the spring stretches 2.76 cm.

Readme [11.4K]

F=K*X,
F=M*a

M*a=K*X

2.5*9.81=K*0.0276

24.525=K*0.0276

24.525/0.0276=K

K= 888.6 N/m ---- force constant

assuming 2.5 refers to the new extension, just divide F/ 0.025
to get

981N/m

8 0

3 years ago

A 4.8 mF capacitor in series with a 500 Ω resistor is connected, by a switch, to a 12 V battery. The current through the resisto

ad-work [718]

Answer:

Current will be 81.7 mA

Which is not given in bellow option

Explanation:

We have given capacitance $C=4.8mF=4.8\times 10^{-3}F$

Resistance R = 500 ohm

Voltage V = 12 volt

We know that time constant of RC circuit of RC circuit is given by

$\tau =RC=500\times 4.8\times 10^{-3}=2.4sec$

Time is given as t = 1 sec

We know that current in RC circuit is given by

$i=\frac{v}{R}(1-e^{\frac{-t}{\tau }})$

So current $i=\frac{12}{500}(1-e^{\frac{-1}{2.4 }})=0.00817A=81.7mA$

Which is not given in the following option

3 0

2 years ago

Read 2 more answers

There is a uniform magnetic field of magnitude B, pervading all space, perpendicular to the plane of rod and rails. The rod is r

Charra [1.4K]

The right hand rule to find the direction of the magnetic field for a falling bar is:

The charge is positive the magnetic field is outgoing, horizontally and towards us.

The charge of the bar is negative, the magnetic field is incoming, that is horizontal away from us.

The magnetic force is given by the vector product of the velocity and the magnetic field.

F = q v x B

Where the bolds indicate vectors, F is the force, q the charge on the particle, v the velocity and B the magnetic field.

In the vector product, the vectors are perpendicular, which is why the right-hand rule has been established, see attached:

The thumb points in the direction of speed.

Fingers extended in the direction of the magnetic field.

The palm is in the direction of the force if the charge is positive and in the opposite direction if the charge is negative.

They indicate that the bar is dropped, therefore its speed is vertical and downwards, it moves to the left therefore this is the direction of the force, we use the right hand rule, the magnetic field must be horizontal, we have two possibilities:

If the charge is positive the magnetic field is outgoing, horizontally and towards us.

If the charge of the bar is negative, the magnetic field is incoming, that is, horizontal away from us

In conclusion using the right hand rule we can find the direction of the magnetic field for a falling bar is:

The charge of the bar is negative, the magnetic field is incoming, that is horizontal away from us.

The charge is positive the magnetic field is outgoing, horizontally and towards us.

Learn more about the right hand rule here: brainly.com/question/12847190

6 0

2 years ago