Answer:
Explanation:
Newton's law of universal gravitation states that the force experimented by a satellite of mass m orbiting Mars, which has mass 
at a distance r will be:
where 
is the gravitational constant.
This force is the centripetal force the satellite experiments, so we can write:
Putting all together:
which means:
![r=\sqrt[3]{\frac{GM}{4\pi^2}T^2}](https://tex.z-dn.net/?f=r%3D%5Csqrt%5B3%5D%7B%5Cfrac%7BGM%7D%7B4%5Cpi%5E2%7DT%5E2%7D)
Which for our values is:
![r=\sqrt[3]{\frac{(6.67\times10^{-11}Nm^2/kg^2)(6.39\times10^{23} kg)}{4\pi^2}(1.026\times24\times60\times60s)^2}=20395282m=20395.3km](https://tex.z-dn.net/?f=r%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%286.67%5Ctimes10%5E%7B-11%7DNm%5E2%2Fkg%5E2%29%286.39%5Ctimes10%5E%7B23%7D%20kg%29%7D%7B4%5Cpi%5E2%7D%281.026%5Ctimes24%5Ctimes60%5Ctimes60s%29%5E2%7D%3D20395282m%3D20395.3km)
Since this distance is measured from the center of Mars, to have the height above the Martian surface we need to substract the radius of Mars R=3389.5 km
, which leaves us with:
From the starting depth to the surface, the vertical distance is 35 ft.
From the surface to the peak of the jump, the vertical distance is 27 ft.
From the peak of the jump to the surface, the vertical distance is 27 ft.
From the surface to the ending depth, the vertical distance is 18 ft.
Then the total vertical distance is ...
35 ft + 27 ft + 27 ft + 18 ft = 107 ft
<h2>The K.E of the charge is 1.02 x 10⁻¹⁷ J</h2>
Explanation:
When the charge of 2e is placed in between the plates .
The force applied on this charge by plates is = q E
here q is the magnitude of charge = 2 e = 2 x 1.6 x 10⁻¹⁹ C
and E is the magnitude of electric field intensity
The work done = Force x displacement
Thus W = q E x S
here S is displacement
Therefore W = 2 x 1.6 x 10⁻¹⁹ x 4 x 8
= 1.02 x 10⁻¹⁷ J
This work will be converted into the kinetic energy of charge .
Thus K.E = 1.02 x 10⁻¹⁷ J
The solution to your problem is as follows:
2.2Kg*9.8m/s = 21.56N
<span>
21.56N*1.25m = 26.95J </span>
<span>We're only concerned with the work done against gravity, lifting the books to 1.25 meters. the distance walked has no effect on the problem, unless you take into account the wind resistance and the force needed to overcome it. Also, lowering the books onto the shelf doesnt count, because gravity does the work on the books.</span>
