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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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B. Find the centripetal force needed to keep an 60.0 kg rider traveling in a circle in the ride.

schepotkina [342]

Answer:

loloololo

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Explanation:

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8 0

2 years ago

Extrusive rocks forms beneath earth's surface true or false

Verizon [17]

The answer is false your welcome

8 0

3 years ago

A baseball thrown by a pitcher is hit by a batter. At the moment when the ball hits the bat, the force exerted on the bat by the

Slav-nsk [51]

Newton's Laws of motion describe the motion of an object based on the applied force

The correct option that gives the relationships between the forces is the the option;

$\underline{F_{ball}}$ on bat = $\underline{F_{bat}}$ on ball

(Either option A or B without the minus symbol before $F_{bat}$ , likely typographical error)

Reason:

Force exerted on the bat by the ball = $F_{ball}$

Force exerted on the ball by the bat = $F_{bat}$

Given that the batter hits the ball with the bat, the force exerted by the bat on the ball, $F_{bat}$ , is reacted to by the force of the ball acting on the bat, $F_{ball}$

According to Newton's third law of motion, action and reaction are equal and opposite. Therefore, at the moment when the ball hits the bat, we have;

$\underline{F_{ball}}$ on bat = $\underline{F_{bat}}$ on ball

Where the force of the bat is high, the ball is accelerated to travel at high speed

Learn more Newton's Laws of motion here:

brainly.com/question/24522313

8 0

3 years ago

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0

3 years ago

Indique onde em quantos metros o rapaz chegará com as seguintes condições: S0 = 0 a) v = 3 m/s e t = 2 s b) v = 2 m/s e t = 3,5

Naily [24]

Answer:

I will answer in English.

Here we will use the relation

Velocity*time = distance

So:

a) velocity = 3m/s

time = 2s

Distance = 3m/s*2s = 6m

b) velocity = 2m/s

time = 3.5s

Distance = 2m/s*3.5s = 7m

c) velocity = 10m/s

time = 0.5s

Distance = 10m/s*0.5s = 5m

d) velocity = 4m/s

time = 2.5s

Distance = 4m/s*2.5s = 9m

e) velocity = 1.5m/s

time = 5s

Distance = 1.5m/s*5s = 7.5m

8 0

3 years ago