Answer:
Open
Explanation:
A switch is a part of a circuit where a connection can be made or broken. By convention, when the switch is "open", the connection is broken and current cannot pass. When the switch is "closed", the connection is complete and current can pass.
The refineries that use the oil to put in their cars as gasoline and then after a while the oil will disappear and go away and that's what a nonrenewable resource would be
When you throw the ball in the air it is considered kinetic energy. Once the ball reaches its max height, it stops moving and all kinetic energy turns into potential energy. So when the ball is at its highest point.
Answer:
Distance = 345719139.4[m]; acceleration = 3.33*10^{19} [m/s^2]
Explanation:
We can solve this problem by using Newton's universal gravitation law.
In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m
![r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]](https://tex.z-dn.net/?f=r_%7Be%7D%20%3D%20distance%20earth%20to%20the%20astronaut%20%5Bm%5D.%5C%5Cr_%7Bm%7D%20%3D%20distance%20moon%20to%20the%20astronaut%20%5Bm%5D%5C%5Cr_%7Bt%7D%20%3D%20total%20distance%20%3D%203.84%2A10%5E8%5Bm%5D)
Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.
Mathematically this equals:

![F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]](https://tex.z-dn.net/?f=F_%7Bm%7D%20%3DG%2A%5Cfrac%7Bm_%7Bm%7D%2Am_%7Ba%7D%20%20%7D%7Br_%7Bm%7D%20%5E%7B2%7D%20%7D%20%5C%5Cwhere%3A%5C%5CG%20%3D%20gravity%20constant%20%3D%206.67%2A10%5E%7B-11%7D%5B%5Cfrac%7BN%2Am%5E%7B2%7D%20%7D%7Bkg%5E%7B2%7D%20%7D%20%5D%20%5C%5Cm_%7Be%7D%3D%20earth%27s%20mass%20%3D%205.98%2A10%5E%7B24%7D%5Bkg%5D%5C%5C%20m_%7Ba%7D%3D%20astronaut%20mass%20%3D%20100%5Bkg%5D%5C%5Cm_%7Bm%7D%3D%20moon%27s%20mass%20%3D%207.36%2A10%5E%7B22%7D%5Bkg%5D)
When we match these equations the masses cancel out as the universal gravitational constant

To solve this equation we have to replace the first equation of related with the distances.

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.
![r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]](https://tex.z-dn.net/?f=r_%7Bm1%2C2%7D%3D%5Cfrac%7B-b%2B-%20%5Csqrt%7Bb%5E%7B2%7D-4%2Aa%2Ac%20%7D%20%20%7D%7B2%2Aa%7D%5C%5C%20%20where%3A%5C%5Ca%3D80.25%5C%5Cb%3D768%2A10%5E%7B6%7D%20%5C%5Cc%20%3D%20-1.47%2A10%5E%7B17%7D%20%5C%5Creplacing%3A%5C%5Cr_%7Bm1%2C2%7D%3D%5Cfrac%7B-768%2A10%5E%7B6%7D%2B-%20%5Csqrt%7B%28768%2A10%5E%7B6%7D%29%5E%7B2%7D-4%2A80.25%2A%28-1.47%2A10%5E%7B17%7D%29%20%7D%20%20%7D%7B2%2A80.25%7D%5C%5C%5C%5Cr_%7Bm1%7D%3D%2038280860.6%5Bm%5D%20%5C%5Cr_%7Bm2%7D%3D-2.97%2A10%5E%7B17%7D%20%5Bm%5D)
We work with positive value
rm = 38280860.6[m] = 38280.86[km]
<u>Second part</u>
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The distance between the Earth and this point is calculated as follows:
re = 3.84 108 - 38280860.6 = 345719139.4[m]
Now the acceleration can be found as follows:
![a = G*\frac{m_{e} }{r_{e} ^{2} } \\a = 6.67*10^{11} *\frac{5.98*10^{24} }{(345.72*10^{6})^{2} } \\a=3.33*10^{19} [m/s^2]](https://tex.z-dn.net/?f=a%20%3D%20G%2A%5Cfrac%7Bm_%7Be%7D%20%7D%7Br_%7Be%7D%20%5E%7B2%7D%20%7D%20%5C%5Ca%20%3D%206.67%2A10%5E%7B11%7D%20%2A%5Cfrac%7B5.98%2A10%5E%7B24%7D%20%7D%7B%28345.72%2A10%5E%7B6%7D%29%5E%7B2%7D%20%20%7D%20%5C%5Ca%3D3.33%2A10%5E%7B19%7D%20%5Bm%2Fs%5E2%5D)