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yKpoI14uk [10]
3 years ago
14

Reena has rolled down her toy car through a ramp. When she increases the height of the ramp (10 cm ) the car has travelled 5 met

res and she increases the height of the ramp (13 cm) the car has travelled 6 metres . What happened when she raises the height of the ramp?
Physics
1 answer:
Stolb23 [73]3 years ago
3 0

Answer:

As she raises the height of the ramp, the speed of the toy car decreases.

Explanation:

Speed is the ratio of the distance traveled by an object to the time taken.

i.e speed = \frac{distance}{time}

From the given question, it would be observed that:

i. when the height of the ramp was 10 cm, the car traveled 5 meters.

ii. when the height of the ramp was increased to 13 cm, the car traveled 6 meters.

Therefore, an increase in the height of the ramp causes a decrease in the speed of the toy car so that the distance traveled per unit time increases. Showing that there is an inverse relationship between the height of the ramp and the speed of the toy car.

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Suppose an asteroid orbiting the sun had an orbital period of 7. 5 years. What would its orbital radius be?.
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By using the orbital period equation we will find that the orbital radius is r = 4.29*10^11 m

<h3>What is the orbital period?</h3>

This would be the time that a given body does a complete revolution in its orbit.

It can be written as:

T = \sqrt{\frac{4*\pi ^2*r^3}{G*M} }

Where:

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Rewriting the equation for the radius we get:

T = \sqrt{\frac{4*\pi ^2*r^3}{G*M} }\\\\r = \sqrt[3]{ \frac{T^2*G*M}{4*\pi ^2} }

Where T = 7.5 years = 7.5*(3.154*10^7 s) = 2.3655*10^8 s

Replacing the values in the equation we get:

r = \sqrt[3]{ \frac{(2.3655*10^8 s)^2*(6.67*10^{-11} m^3/(kg*s^2))*(1.989*10^{30} kg)}{4*3.14 ^2} } = 4.29*10^{11 }m

So the orbital radius is 4.29*10^11 m

If you want to learn more about orbits, you can read:

brainly.com/question/11996385

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