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Inessa05 [86]
3 years ago
11

The Moon takes about 27 days to orbit the Earth. Assuming a circular orbit, how fast is it orbiting? Express your answer in km/h

r.
Physics
1 answer:
Vinvika [58]3 years ago
7 0

Answer:

3727.24km/h

Explanation:

Hello!

To solve this problem we must first know what is the distance from the earth to the moon, this will be our radius.

=384400Km

Then we find the distance traveled which would be the perimeter of a circle = 2πr, finally to find the speed we divide the distance traveled by 27 days.

Finally we use conversion factor to have the speed in km / h

solving

perimeter= 2(3.14)r=2(3.14)(384400Km)=2415256.432km\\Speed=\frac{2415256.432km}{27days} *\frac{1day}{24h} =3727.24km/h

the moon is orbiting at speed of 3727.24km/h

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The force of friction is 213.2 N

Explanation:

The frictional force acting on an object sliding on a surface is given by:

F_f = \mu mg

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m is the mass of the object

g is the acceleration of gravity

In this problem we have

\mu=0.750

m = 29 kg

g=9.8 m/s^2

Therefore, the force of friction is

F_f = (0.750)(29)(9.8)=213.2 N

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3 years ago
In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of t
muminat
<h2><em><u>⇒</u></em>Answer:</h2>

In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, g . How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)

Step-by-Step Solution:

Solution 35PE

This question discusses about the increased range. So, we shall assume that the angle of jumping will be  as the horizontal range is maximum at this angle.

Step 1 of 3<

/p>

The legs have an extension of 0.600 m in the crouch position.

So,  m

The person is at rest initially, so the initial velocity will be zero.

The acceleration is  m/s2

Acceleration  m/s2

Let the final velocity be .

Step 2 of 3<

/p>

Substitute the above given values in the kinematic equation  ,

m/s

Therefore, the final velocity or jumping speed is  m/s

Explanation:

3 0
3 years ago
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\\ \bull\tt\longmapsto \lambda=\dfrac{c}{v}

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It has unit as meter(m)

8 0
3 years ago
A wheel with moment of inertia 25 kg. m2 and angular velocity 10 rad/s begins to speed up, with angular acceleration 15 rad/sec2
Pani-rosa [81]

Answer:

(A) Angular speed 40 rad/sec

Rotation = 50 rad

(b) 37812.5 J

Explanation:

We have given moment of inertia of the wheel I=25kgm^2

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Angular acceleration \alpha =15rad/sec^2

(a) We know that \omega =\omega _0+\alpha t

We have given t = 2 sec

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Now \Theta =\omega _0t+\frac{1}{2}\alpha t^2=10\times 2+\frac{1}{2}\times 15\times 2^2=50rad

(b) After 3 sec \omega =10+15\times 3=55rad/sec

We know that kinetic energy is given by Ke=\frac{1}{2}I\omega ^2=\frac{1}{2}\times 25\times 55^2=37812.5J

7 0
3 years ago
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Answer:

Solid gold. 10 carat indicates gold purity is not 100%. For 24 karat gold is pure gold.

Explanation:

8 0
3 years ago
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