Answer:
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Answer:
Tension of 132N
Explanation:
We need to apply Summatory of Force to find the tension in the hand.
We define te tensión in the hand as
and the Tension in fence post as
, then
![\sum F = 24(9.8)](https://tex.z-dn.net/?f=%5Csum%20F%20%3D%2024%289.8%29)
![F_1 + F_2= 24(9.8)](https://tex.z-dn.net/?f=F_1%20%2B%20F_2%3D%2024%289.8%29)
We apply summatory of moments then
![F_2*1.25 = F_1*1.6](https://tex.z-dn.net/?f=F_2%2A1.25%20%3D%20F_1%2A1.6)
Where the Force 2 is 1.25m from the center of summatory,
We can note that,
![1.6 m - 0.35m=1.25m](https://tex.z-dn.net/?f=1.6%20m%20-%200.35m%3D1.25m)
We have two equation and two incognites, then replacing (1) in (2)
![1.6(235.2 -F_2) = 1.25F_2](https://tex.z-dn.net/?f=1.6%28235.2%20-F_2%29%20%3D%201.25F_2)
![376.32 = F2(1.6+1.25)](https://tex.z-dn.net/?f=376.32%20%3D%20F2%281.6%2B1.25%29)
![F_2= \frac{376.32}{2.85}](https://tex.z-dn.net/?f=F_2%3D%20%5Cfrac%7B376.32%7D%7B2.85%7D)
![F_2 =132 N](https://tex.z-dn.net/?f=F_2%20%3D132%20N)
It would be the first one and the third one
<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
Answer:
Greatest gravitational energy is at "C".
The planet has to do work "against" the field to get to "C".
Also, if m v R (angular momentum) is constant then as R increases v must decrease for this term to be constant and KE = 1/2 M v^2 must decrease also to get to point C.