A what is any small object that orbits a larger object
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Answer:
a)
Explanation:
- A block sliding down an inclined plane, is subject to two external forces along the slide.
- One is the component of gravity (the weight) parallel to the incline.
- If the inclined plane makes an angle θ with the horizontal, this component (projection of the downward gravity along the incline, can be written as follows:
(taking as positive the direction of the movement of the block)
- The other force, is the friction force, that adopts any value needed to meet the Newton's 2nd Law.
- When θ is so large, than the block moves downward along the incline, the friction force can be expressed as follows:
- The normal force, adopts the value needed to prevent any vertical movement through the surface of the incline:
- In equilibrium, both forces, as defined in (1), (2) and (3) must be equal in magnitude, as follows:
- As the block is moving, if the net force is 0, according to Newton's 2nd Law, the block must be moving at constant speed.
- In this condition, the friction coefficient is the kinetic one (μk), which can be calculated as follows:
The work required to pump out the liquid is 2.35 MJ.
What is work?
Work can be defined as the scalar product of the force and the displacement.
The formula to calculate the work done W is,
where F is the force and d is the displacement.
Note: It is assumed that the weight of one cubic unit of the liquid is .
Break down the vertical tank into parallel disc elements whose thickness is dh and are located at a height h from the ground. The radius of the element at height h is r.
It is given that the height of the tank is 6 m and the base radius is 5 meters. Using these, the ratio of the radius r at height h to the height h by using the triangle similarity criterion is,
Using the above value of r, the volume of the element is given by,
The weight of the volume element is the force required.
The weight of one cubic unit of the liquid is .
So the weight F of the volume element at height h is given by,
At the height h, the difference in the height of the top of the tank and the volume element is (6-h). So the work required to pump out the volume of liquid at height h is given by the integrating the equation,
The value of the W obtained is 2.35*10^6 J or 2.35 MJ.
Learn more about work here:
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