Answer:
the first one stationary charge
Power = (energy) / (time)
= (1370 joules) / (100 seconds)
= 13.7 joules/second
= 13.7 watts .
That's not an awful lot of power, especially for a strenuous activity like
rock-climbing. Shoot ! Even I could probably perform at that level.
Compare 13.7 watts to the light power coming out of a 20-watt night light.
13.7 watts = 0.018 horsepower. (rounded)
Answer:
P₂ = 1.22 kPa
Explanation:
This problem can be solved using the equation of state:

where,
P₁ = initial pressure = 1 KPa
P₂ = final pressure = ?
V₁ = initial Volume = 1 liter
V₂ = final volume = 1.1 liter
T₁ = initial temperature = 290 k
T₂ = final temperature = 390 k
Therefore,

<u>P₂ = 1.22 kPa</u>
Hello. You did not inform the experiment that Arthur is conducting, which makes it impossible for your question to be answered accurately. However, I will try to help you in the best possible way.
The hypothesis is an assumption that is made before the experiment is carried out. This hypothesis is formed with the observation of some phenomenon of nature where the researcher believes that two or more elements interact to form a result. In this case, the experiment is carried out to determine whether the assumption, that is, the hypothesis is false or true. In the event that an experiment determines that the hypothesis is false, two things may have occurred: (a) the experiment was set up, or analyzed incorrectly, (b) the elements tested have no relation to the observed phenomenon.
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>
<u />
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)