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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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Please help with my geology hw

Dominik [7]

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3 years ago

The cylindrical tub of a dryer in a laundromat rotates counterclockwise about a horizontal axis at 41.5 rev/min as it dries the

frozen [14]

Answer:

$\theta = 49.81^0$

Explanation:

Given that:

$\omega = 41.5 \ rev/min\\\\\omega = 41.5 *\frac{1}{60}* 2 \pi\\\\\omega = 4.45 \ rad/s\\\\\\diameter = 0.748 m$

If we let the piece of the close lose contact at ∠θ;

Then ; from force balance;

we have:

$\\\\mg sin \theta = \frac{mv^2}{r}\\\\sin \theta = \frac{2v^2}{dg}\\\\\theta = sin^{-1} (\frac{2v^2}{dg})$

where;

$v = \frac{\omega d}{2}\\\\v = \frac{4.45 *0.748}{2}\\\\v = 1.6643\\\\v^2 = 2.77$

Again:

$\theta = sin^{-1}(\frac{2v^2}{dg})\\\\\theta = sin^{-1}( \frac{2*2.77}{0.74*9.8})\\\\\theta = 49.81^0$

6 0

3 years ago

As you move from left to right across a period the number of valances electrons

oksian1 [2.3K]

The valence electrons increase as you move left to right.

3 0

3 years ago

Read 2 more answers

The current world-record motorcycle jump is 77.0 m, set by Jason Renie. Assume that he left the take-off ramp at 13.0° to the ho

kotegsom [21]

Answer:

The take-off speed is 41.48 $\frac{m}{s}$

Explanation:

Given :

Range $R = 77$ m

Projectile angle $\alpha =$ 13°

From the formula of range,

$R = \frac{v_{o} ^{2} }{g} \sin 2\alpha$

Find the velocity from above equation,

$v_{o} = \sqrt{\frac{gR}{\sin 2 \alpha } }$

$v_{o} = \sqrt{\frac{9.8 \times 77}{\sin 26} }$ ( ∵ $g = 9.8$ $\frac{m}{s^{2} }$ )

$v_{o} = 41.48$ $\frac{m}{s}$

Therefore, the take-off speed is 41.48 $\frac{m}{s}$

7 0

4 years ago

You are given three pieces of wire that have different shapes (dimensions). You connect each piece of wire separately to a batte

VikaD [51]

Answer:

$R_3 < R_1 < R_2$

Explanation:

The resistance of a wire is given by:

$R=\frac{\rho L}{A}$

where

$\rho$ is the resistivity of the material

L is the length of the wire

A is the cross-sectional area of the wire

1) The first wire has length L and cross-sectional area A. So, its resistance is:

$R_1=\frac{\rho L}{A}$

2) The second wire has length twice the first one: 2L, and same thickness, A. So its resistance is

$R_2=\frac{2\rho L}{A}$

3) The third wire has length L (as the first one), but twice cross sectional area, 2A. So, its resistance is

$R_3=\frac{\rho L}{2A}$

By comparing the three expressions, we find

$R_3 < R_1 < R_2$

So, this is the ranking of the wire from most current (least resistance) to least current (most resistance).

5 0

3 years ago