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CaHeK987 [17]
3 years ago
6

Your friend says that the force that the sun exerts on earth is much larger than the force that earth exerts on the sun. part a

do you agree or disagree with this opinion? do you agree or disagree with this opinion? agree disagree
Physics
2 answers:
Verizon [17]3 years ago
8 0
I disagree with that opinion, and I have solid Physics to back me up.

The forces of gravity are always equal in both directions. The sun pulls the Earth with exactly the same force with which the Earth pulls the sun.

It may seem weird, but your weight on Earth is exactly the same as the Earth's weight on you. For the same reason.
maria [59]3 years ago
8 0

Answer:

<h2>Complete disagree.</h2>

Explanation:

The gravity force, according to Newton, is defined as

F_{g}=G\frac{m_{1}m_{2}}{r_{1,2} ^{2} }

This definition expresses that a gravitational force is formed by a system of to masses to bodies, the more mass, more force. However, this definition also states that the gravitational force is not different for each mass that is involved. That is, the gravitational force on Earth is the same than the one exerted by the Sun, because such force is formed by the system Sun-Earth.

In addition, the gravitational force is inversely proportional to the distance between the bodies, that is, the more distance, less gravitational force, the less distance, less gravitational force.

Therefore, the right answer here is a complete disagree.

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A wheel that is rotating at 33.3 rad/s is given an angular acceleration of 2.15 rad/s 2. Through what angle has the wheel turned
Neporo4naja [7]

Answer:

The angle through which the wheel turned is 947.7 rad.

Explanation:

initial angular velocity, \omega _i = 33.3 rad/s

angular acceleration, α = 2.15 rad/s²

final angular velocity, \omega_f = 72 rad/s

angle the wheel turned, θ = ?

The angle through which the wheel turned can be calculated by applying the following kinematic equation;

\omega_f^2 = \omega_i^2 + 2\alpha \theta\\\\\theta = \frac{\omega_f^2\   -\  \omega_i^2}{2\alpha } \\\\\theta = \frac{(72)^2\   -\  (33.3)^2}{2(2.15)}\\\\\theta = 947.7 \ rad

Therefore, the angle through which the wheel turned is 947.7 rad.

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Please see the attached file.

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