Answer:
(3) The period of the satellite is independent of its mass, an increase in the mass of the satellite will not affect its period around the Earth.
(4) he gravitational force between the Sun and Neptune is 6.75 x 10²⁰ N
Explanation:
(3) The period of a satellite is given as;
![T = 2\pi \sqrt{\frac{r^3}{GM} }](https://tex.z-dn.net/?f=T%20%3D%202%5Cpi%20%5Csqrt%7B%5Cfrac%7Br%5E3%7D%7BGM%7D%20%7D)
where;
T is the period of the satellite
M is mass of Earth
r is the radius of the orbit
Thus, the period of the satellite is independent of its mass, an increase in the mass of the satellite will not affect its period around the Earth.
(4)
Given;
mass of the ball, m₁ = 1.99 x 10⁴⁰ kg
mass of Neptune, m₂ = 1.03 x 10²⁶ kg
mass of Sun, m₃ = 1.99 x 10³⁰ kg
distance between the Sun and Neptune, r = 4.5 x 10¹² m
The gravitational force between the Sun and Neptune is calculated as;
![F_g = \frac{Gm_2m_3}{r^2} \\\\F_g = \frac{6.67\times 10^{-11} \times 1.03 \times 10^{26}\times 1.99\times 10^{30}}{(4.5\times 10^{12})^2} \\\\F_g = 6.751 \times 10^{20} \ N](https://tex.z-dn.net/?f=F_g%20%3D%20%5Cfrac%7BGm_2m_3%7D%7Br%5E2%7D%20%5C%5C%5C%5CF_g%20%3D%20%5Cfrac%7B6.67%5Ctimes%2010%5E%7B-11%7D%20%5Ctimes%201.03%20%5Ctimes%2010%5E%7B26%7D%5Ctimes%201.99%5Ctimes%2010%5E%7B30%7D%7D%7B%284.5%5Ctimes%2010%5E%7B12%7D%29%5E2%7D%20%5C%5C%5C%5CF_g%20%3D%206.751%20%5Ctimes%2010%5E%7B20%7D%20%5C%20N)
Answer:
![v = 15.45 m/s](https://tex.z-dn.net/?f=v%20%3D%2015.45%20m%2Fs)
Explanation:
As per mechanical energy conservation we can say that here since friction is present in the barrel so we will have
Work done by friction force = Loss in mechanical energy
so we will have
![W_f = (U_i + K_i) - (U_f + K_f)](https://tex.z-dn.net/?f=W_f%20%3D%20%28U_i%20%2B%20K_i%29%20-%20%28U_f%20%2B%20K_f%29)
here we know that
![W_f = F_f . d](https://tex.z-dn.net/?f=W_f%20%3D%20F_f%20.%20d)
![W_f = 40 \times 4](https://tex.z-dn.net/?f=W_f%20%3D%2040%20%5Ctimes%204)
![W_f = 160 J](https://tex.z-dn.net/?f=W_f%20%3D%20160%20J)
Initial compression in the spring is given as
![F = kx](https://tex.z-dn.net/?f=F%20%3D%20kx)
![4400 = 1100 x](https://tex.z-dn.net/?f=4400%20%3D%201100%20x)
![x = 4 m](https://tex.z-dn.net/?f=x%20%3D%204%20m)
now from above equation
![W_f = (\frac{1}{2}kx^2 + 0) - (mgh + \frac{1}{2}mv^2)](https://tex.z-dn.net/?f=W_f%20%3D%20%28%5Cfrac%7B1%7D%7B2%7Dkx%5E2%20%2B%200%29%20-%20%28mgh%20%2B%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2%29)
![160 = (\frac{1}{2}1100(4^2) + 0) - (60 \times 9.8\times 2.50 + \frac{1}{2}(60)v^2)](https://tex.z-dn.net/?f=160%20%3D%20%28%5Cfrac%7B1%7D%7B2%7D1100%284%5E2%29%20%2B%200%29%20-%20%2860%20%5Ctimes%209.8%5Ctimes%202.50%20%2B%20%5Cfrac%7B1%7D%7B2%7D%2860%29v%5E2%29)
![160 = 8800 - 1470 - 30 v^2](https://tex.z-dn.net/?f=160%20%3D%208800%20-%201470%20-%2030%20v%5E2)
![v = 15.45 m/s](https://tex.z-dn.net/?f=v%20%3D%2015.45%20m%2Fs)
To find work, you use the equation: W = Force X Distance X Cos (0 degrees)
Following the Law of Conservation of Energy, energy cannot be destroyed nor created.
So you would do 75 N x 10m x Cos (0 degrees)= 750 J
Answer:
the more particles packed together the faster it falls
Explanation:
the mass + the 1 constant g-force = the speed without adding air resistance
The freezing point is the same as the melting point.
If it freezes at -58°C, hence the melting point is also <span>-58°C.</span>