Let's think about the system that includes the spacecraft, the astronaut,
and all the equipment tied to him. In this system, nothing is moving relative
to anything else, so the total linear momentum is zero.
Now, suddenly, a blast of gas leaves the astronaut's little SCUBA tank, and
hisses away from him and the spacecraft, in that direction ===> .
The momentum of the cloud of nitrogen is
(mass) x (speed)
= (0.03 kg) x (900 m/s)
= 27 kg-m/s in that direction ===> .
Now is the perfect time to recall that momentum is conserved.
It can't be suddenly created or destroyed, it can't appear out of
nowhere, it can't disappear into nowhere, and the total amount
of momentum in a system is constant.
In order to keep the total momentum of this system constant, the
astronaut himself gets 27 kg-m/s of momentum in this direction <=== .
The mass of him and his equipment is 320 kg.
So ...
27 kg-m/s <== = (320 kg) x (speed <==)
Speed = 27 kg-m/s / 320 kg
= 0.084375 m/s in this direction <===
Answer:
![V = 1.44\times 10^{5}~V](https://tex.z-dn.net/?f=V%20%3D%201.44%5Ctimes%2010%5E%7B5%7D~V)
Explanation:
The electric potential can be found by using the following formula
![V = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cfrac%7BQ%7D%7Br%5E2%7D)
Applying this formula to each charge gives the total potential.
![V = V_1 + V_2 + V_3 + V_4\\V = \frac{1}{4\pi\epsilon_0}\frac{4\times 10^{-6}}{(\sqrt{2}/2)^2} + \frac{1}{4\pi\epsilon_0}\frac{4\times 10^{-6}}{(\sqrt{2}/2)^2} + \frac{1}{4\pi\epsilon_0}\frac{3\times 10^{-6}}{(\sqrt{2}/2)^2} - \frac{1}{4\pi\epsilon_0}\frac{3\times 10^{-6}}{(\sqrt{2}/2)^2}\\V = \frac{16\times 10^{-6}}{4\pi\epsilon_0}\\V = 1.44\times 10^{5}~V](https://tex.z-dn.net/?f=V%20%3D%20V_1%20%2B%20V_2%20%2B%20V_3%20%2B%20V_4%5C%5CV%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cfrac%7B4%5Ctimes%2010%5E%7B-6%7D%7D%7B%28%5Csqrt%7B2%7D%2F2%29%5E2%7D%20%2B%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cfrac%7B4%5Ctimes%2010%5E%7B-6%7D%7D%7B%28%5Csqrt%7B2%7D%2F2%29%5E2%7D%20%2B%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cfrac%7B3%5Ctimes%2010%5E%7B-6%7D%7D%7B%28%5Csqrt%7B2%7D%2F2%29%5E2%7D%20-%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cfrac%7B3%5Ctimes%2010%5E%7B-6%7D%7D%7B%28%5Csqrt%7B2%7D%2F2%29%5E2%7D%5C%5CV%20%3D%20%5Cfrac%7B16%5Ctimes%2010%5E%7B-6%7D%7D%7B4%5Cpi%5Cepsilon_0%7D%5C%5CV%20%3D%201.44%5Ctimes%2010%5E%7B5%7D~V)
Since the potential is a scalar quantity, it is safe to sum all the potentials straightforward. And since they all placed on the corners of a square, +3 and -3 μC charges cancel out each other.
Answers:
a) ![8.009(10)^{-7} N](https://tex.z-dn.net/?f=8.009%2810%29%5E%7B-7%7D%20N)
b) ![1.8(10)^{-8}](https://tex.z-dn.net/?f=1.8%2810%29%5E%7B-8%7D%20)
Explanation:
a) Accoding to the Universal Law of Gravitation we have:
(1)
Where:
is the gravitational force between the eagle and the throng
is the Universal Gravitational constant
is the mass of the eagle
is the mass of the throng
is the distance between the throng and the eagle
(2)
(3) As we can see the gravitational force between the eagle and the throng is quite small.
b) The attraction force between the eagle and Earth is the weight
of the eagle, which is given by:
(4)
Where
is the acceleration due gravity on Earth
(5)
(6)
Now we can find the ratio between
and
:
![\frac{F_{g}}{W}=\frac{8.009 (10)^{-7} N}{44.1N}](https://tex.z-dn.net/?f=%5Cfrac%7BF_%7Bg%7D%7D%7BW%7D%3D%5Cfrac%7B8.009%20%2810%29%5E%7B-7%7D%20N%7D%7B44.1N%7D)
As we can see this ratio is also quite small