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ivanzaharov [21]
3 years ago
8

How many days does the earth take to rotate around the sun

Physics
1 answer:
ValentinkaMS [17]3 years ago
3 0

Answer:

1 day

Explanation:

The spinning movement of Earth around its axis is known as Earth's rotation.

The Earth rotates around the Sun in 1 day that is 24 hours but once in every 23 hours, 56 minutes, and 4 seconds with respect to other stars.

According to the Indian astronomer Aryabhata, Earth rotates about its axis daily.

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A 35 kg trunk is pushed 4.8 m at constant speed up a 30° incline by a constant horizontal force. The coefficient of kinetic fric
netineya [11]

Answer:

We need to separate the x- and y-components of the applied force. For simplicity, I will denote the direction along the inclined plane as x-direction, and the perpendicular direction as y-direction.

F_x = F\cos(30^\circ)\\F_y = F\sin(30^\circ)

Only the x-component of the applied horizontal force does work on the trunk.

But we need to find the magnitude of the force. We know that the trunk is moving with constant speed. So, the x-component of the applied force is equal to the x-component of the gravitational force plus the force of friction.

0 = F\cos(30^\circ) - mg\sin(30^\circ) - mg\cos(30^\circ)\mu\\0.86F = 35(9.8)(0.5) + 35(9.8)(0.86)(0.19)\\F =264.5N

F_x = 264.5\cos(30^\circ) = 227.5N\\F_y = 264.5\sin(30^\circ) = 132.25N

W_{F_x} = F_xd = 227.5\times 4.8 = 1092J

The work done by the weight of the trunk can be calculated similarly. Only the x-component of the weight does work on the trunk.

W_g = -mg\sin(30^\circ)d

Note that the direction of the weight force is opposite of the direction of the motion, so this force does negative work on the trunk.

W_g = -823J

The energy dissipated by the frictional force can be found as follows:

W_f = -mg\cos(30^\circ)\mu d = -35(9.8)(0.86)(0.19)(4.8) = -269J

Additionally, the sum of work done by the friction and weight is equal in magnitude to the work done by the applied force. This shows that our calculations are consistent.

In the second part of the question, the applied force is on the x-direction. We will follow a similar procedure but a different force.

0 = F - mg\sin(30^\circ) - mg\cos(30^\circ)\mu\\0 = F - 35(9.8)(0.5) - 35(9.8)(0.86)(0.19)\\0 = F - 171.5 - 56\\F = 227.5N

W_F = Fd = 227.5\times 4.8 = 1092J

W_g = -mg\sin(30^\circ)d = -35(9.8)(0.5)(4.8) = -823J

W_f = -mg\cos(30^\circ)\mu d = -35(9.8)(0.86)(0.19)(4.8) = -269J

Explanation:

As you can see above, the answers are the same, although the directions of the applied forces are different. The reason for this situation is that in the first part the y-component does no work.

8 0
4 years ago
____ have more energy than ____.
Katena32 [7]
Ultraviolet rays; X-rays
8 0
4 years ago
Read 2 more answers
The half-lives for the radioactive decay of various elements are listed below.
myrzilka [38]

Answer:

D ko alam pasensya ka na ha

4 0
3 years ago
A car enters a freeway with initial velocity of 15.0 m/s and with con stant rate of acceleration, reaches a velocity of 22.5 m/s
LiRa [457]

Answer:

a) The acceleration is 2.14 m/s^{2}

b) The distance traveled by the car is 65.61 m

c) The average velocity is 18.75 m/s

Explanation:

Using the equations that describe an uniformly accelerated motion:

a) a=\frac{v_f - v_o}{t} =\frac{22.5m/s - 15.0 m/s}{3.50s}

b) d= d_0 + v_0 t + \frac{1}{2} a t^{2} = 0 +15.0 x 3.5 + \frac{2.14x3.50^{2} }{2} = 65.61 m

c) v_m =\frac{d}{t}=\frac{65.61}{3.5}  =18.75 m/s

8 0
4 years ago
An ostrich can run at speeds of up to 72 kilometers per hour. How long will it take an ostrich to run 1.5 kilometers at this top
tatiyna
S=d/t
T=d/s
= 1.5/72
= 0.02 hours
6 0
4 years ago
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