The directions arrow<span> is </span>always<span> going the wrong </span>way<span>.</span>
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Answer:
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The velocity of a satellite describing a circular orbit is <u>constant</u> and defined by the following expression:
(1)
Where:
is the gravity constant
the mass of the massive body around which the satellite is orbiting
the radius of the orbit (measured from the center of the planet to the satellite).
Note this orbital speed, as well as orbital period, does not depend on the mass of the satellite. I<u>t depends on the mass of the massive body.</u>
In addition, this orbital speed is constant because at all times <u>both the kinetic energy and the potential remain constant</u> in a circular (closed) orbit.
Let us consider body moves a distance S due to the force F.
Hence the work by the body W = FS
If the force is not along the direction of displacement,then the work by a body for travelling a distance S will be -
where
is the component of the force along the direction of displacement.


As per the question the power P is given as -




Hence alternative definition of power P = F.V
For a photographer that wishes to determine the color of light that he can use in a dark room that will not expose the films he is processing, having used a Blue Incandescent bulb, he should proceed to use a Red Incandescent bulb for the next trial.
The photographer in question is performing an experiment. For these kinds of experiments it is important to identify the variables present, which can be of three kinds:
- Control variables
- Dependent variables
- Independent variables
For this experiment, the dependent variable is the exposure of the light onto the films, given that this is what we wish to measure. The independent variable will be the color of the light being used which is what will affect the dependent variable.
The remaining variable must be the control variable. Unlike the previous variables, we can have more than one of these. The control variable is there to make sure that only the dependent variable is affecting the outcome. We do this by keeping the control variable the same through each trial, which is why the photographer should not change the type of bulb in the second experiment, changing only the color of the light.
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