In a particular machine, there are 2 gears that interlock; One gear is larger in circumference than the other. The manufacturer
1 answer:
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In a particular machine, there are 2 gears that interlock; One gear is larger in circumference than the other. The manufacturer of the gears guarantees that each gear will last for at least 6,000,000,000 revolutions. Assuming that there is no slippage between the 2 gears and that when one gear rotates the other gear also rotates, the larger gear is guaranteed to last how many days longer than the smaller gear?
(1) The diameter of the larger gear is twice the diameter of smaller gear.
(2) The smaller gear revolves 600 times per minute.
Answer:
Number of days larger gear last longer than small gear=115.74 days
Explanation:
Given Data
Revolution=6,000,000,000 revolutions
diameter of larger gear is twice diameter of smaller gear
Smaller gear revolves=600 times per minute
Number of days larger gear last longer than small gear=?
Solution
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<h2>Answer: 56.718 min</h2>
Explanation:
According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>
In other words, this law states a relation between the orbital period of a body (moon, planet, satellite) orbiting a greater body in space with the size
of its orbit.
This Law is originally expressed as follows:
(1)
Where;
is the Gravitational Constant and its value is
is the mass of Mars
is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)
If we want to find the period, we have to express equation (1) as written below and substitute all the values:
(2)
(3)
(4)
Finally:
This is the orbital period of a spacecraft in a low orbit near the surface of mars