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Leno4ka [110]
3 years ago
14

What is the centripetal acceleration of the earth as it travels around the sun when Earth has an orbital radius of 1.5 x 10^11 m

and an average orbital speed of 29.7 km/s
Physics
1 answer:
masya89 [10]3 years ago
4 0

Answer: 0.0058 m/s^{2}

Explanation:

Centripetal acceleration a_{C} is calculated by the following equation:

a_{C}=\frac{V^{2}}{r}

Where:

V=29.7 \frac{km}{h} \frac{1000 m}{1 km}=29700 m/s is the Earth's orbital speed

r=1.5(10)^{11} m is the orbital radius

a_{C}=\frac{(29700 m/s)^{2}}{1.5(10)^{11} m}

a_{C}=0.0058 m/s^{2}

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A 150kg motorcycle starts from rest and accelerates at a constant rate along a distance of 350m. The applied force is 250N and t
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A) The net force on the motorbike is 205.9 N

B) The acceleration of the motorbike is 1.37 m/s^2

C) The final speed is 5.2 m/s

D) The elapsed time is 3.80 s

Explanation:

A)

There are two forces acting on the motorbike:

- The applied force, F = 250 N, forward

- The frictional force, F_f, backward

The frictional force can be written as

F_f = \mu mg

where

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Therefore the net force is given by

\sum F = F - F_f = F - \mu mg

And substituting, we find

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2)

The acceleration of the motorbike can be found by using Newton's second law, which states that the net force is equal to the product between mass and acceleration:

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a is the acceleration

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\sum F = 205.9 N is the net force

m = 150 kg is the mass

Solving for a, we find the acceleration:

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C)

Since the motion of the motorbike is a uniformly accelerated motion, we can use the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For this motorbike, we have:

u = 0 (it starts from rest)

a=1.37 m/s^2

s = 350 m

Solving for v,

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4)

For this part of the problem, we can use the following suvat equation:

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a is the acceleration

t is the elapsed time

Here we have:

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Solving for t, we find

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