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3 years ago

How is constant acceleration indicated on a motion map?

Physics

2 answers:

sasho [114]3 years ago

6 0

<span>by vectors that are all the same length

</span>

saw5 [17]3 years ago

6 0

Answer;

By vectors that are all the same length

Explanation;

A motion map represents the position, velocity, and acceleration of an object at various clock readings.The objects's velocity is represented by a vector.

Constant Acceleration is when the velocity of an object changes its speed uniformly.

Motion with constant acceleration occurs in everyday life whenever an object is dropped: the object moves downward with the constant acceleration 9.81 m/s², under the influence of gravity.

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How would the speed of Earth's orbit around the sun change if Earth's distance from the sun

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If Earth’s distance from the sun increased by 4 times then the speed of Earth’s orbit around the sun would decrease by a factor of 2.

Answer: Option 3

<u>Explanation:</u>

From Newton’s second law of motion, it is known that any kind of external force acting on an object will be equal to object’s mass and acceleration exerted on the object. So in this case, the gravitational force between Earth and Sun will be equal to the mass (M) and acceleration exerted on Earth.

$\text {Gravitational force}=\text {M} \times \text {Acceleration of Earth around the orbit}$

It is known that,

$\text {Acceleration in orbital motion}=\frac{\left(\text {velocity at which the Earth rotates)}^{2}\right.}{\text { Distance of Earth from the Sun}}$

Thus, substituting this, we get,

$\text {Gravitational force}=\frac{\text {M} \times\left(\text {velocity at which the Earth rotates)}^{2}\right.}{\text { Distance of Earth from the sun}}$

Also, we know,

$\text {Gravitational force}=\frac{G \times M \text { of sun } \times \text {M of Earth}}{(\text {Distance of Earth from the Sun})^{2}}$

Then, comparing both the equation, we get the orbital velocity as,

$\text { Orbital velocity } v=\sqrt{\frac{G \times M \text { of Sun }}{\text { Distance of Earth from the Sun }}}=\sqrt{\frac{G M}{r}}$

So, here G is gravitational constant, M is the mass of Sun and r is the distance of separation of Earth from Sun.

If the distance of Earth from Sun increases by 4 times so r’ will be 4r. Thus the new orbital velocity v’ will be

$v^{\prime}=\sqrt{\frac{G M}{r^{\prime}}}=\sqrt{\frac{G M}{4 r}}=\frac{1}{2} \sqrt{\frac{G M}{r}}$

So,

$v^{\prime}=\frac{1}{2} v$

Thus, the orbital speed will be decreased by a factor of 2 when the distance of Earth from the Sun increased by 4 times.

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