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)
The answer is A. Or the first option. Pressure is changed by lowering the pressure, not reducing the volume. You would assume its C but its A.
Answer:
Chicken and peanut butter lol
Explanation:
Answer:
The vulture loses 6.1 m height
Explanation:
Please see the attached figure.
The horizontal distance and the loss of height form a 90º triangle.
The loss of height is the side opposite the given angle (3.5º) and the 100 m horizontal distance is adjacent the angle.
Then, using trigonometric rules:
(1) sin 3.5º = h / hyp
(2) cos 3.5º = distance / hyp
where
h = height lost during the flight.
hyp = hypotenuse of the triangle
Using (2) we can calculate the hypotenuse:
cos 3.5º = 100 m / hyp
hyp = 100 m / cos 3.5º = 100.2 m
with the hypotenuse we can now calculate the loss of height using (1):
sin 3.5º = h / hyp
sin 3.5º = h / 100.2 m
sin 3.5º * 100.2 m = h
<u>h = 6.1 m</u>
( very modest drop in height indeed!)
Answer:
Explanation:
Look at the equation for Potential Energy. PE = mass times gravity times the height. Filling in and solving for h:
34.3 = .5(9.8)h so
34.3 = 4.9h so
h = 7 meters