Sun is the main source of energy
Answer:
A. 70 m OS the correct one
So as you may know atoms are neutral because the number of protons (+ charge) and the number of electrons( - charge) are the same so they cancel out. When a valence electron leaves an atom it will have a +1 charge because there is one less negative than positives or there is one more positive than negatives since a negative electron left. If a valence electron is added a -1 charge because there is now one more negative than positive!!!
hope that helps!!
V = u + at
0 = u -9.8 x 5.5
u = 9.8 x 5.5 = 53.9 m/s
Answer:
![1.69\cdot 10^{10}J](https://tex.z-dn.net/?f=1.69%5Ccdot%2010%5E%7B10%7DJ)
Explanation:
The total energy of the satellite when it is still in orbit is given by the formula
![E=-G\frac{mM}{2r}](https://tex.z-dn.net/?f=E%3D-G%5Cfrac%7BmM%7D%7B2r%7D)
where
G is the gravitational constant
m = 525 kg is the mass of the satellite
is the Earth's mass
r is the distance of the satellite from the Earth's center, so it is the sum of the Earth's radius and the altitude of the satellite:
![r=R+h=6370 km +575 km=6945 km=6.95\cdot 10^6 m](https://tex.z-dn.net/?f=r%3DR%2Bh%3D6370%20km%20%2B575%20km%3D6945%20km%3D6.95%5Ccdot%2010%5E6%20m)
So the initial total energy is
![E_i=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24} kg)}{2(6.95\cdot 10^6 m)}=-1.51\cdot 10^{10}J](https://tex.z-dn.net/?f=E_i%3D-%286.67%5Ccdot%2010%5E%7B-11%7D%29%5Cfrac%7B%28525%20kg%29%285.98%5Ccdot%2010%5E%7B24%7D%20kg%29%7D%7B2%286.95%5Ccdot%2010%5E6%20m%29%7D%3D-1.51%5Ccdot%2010%5E%7B10%7DJ)
When the satellite hits the ground, it is now on Earth's surface, so
![r=R=6370 km=6.37\cdot 10^6 m](https://tex.z-dn.net/?f=r%3DR%3D6370%20km%3D6.37%5Ccdot%2010%5E6%20m)
so its gravitational potential energy is
![U = -G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24}kg)}{6.37\cdot 10^6 m}=-3.29\cdot 10^{10} J](https://tex.z-dn.net/?f=U%20%3D%20-G%5Cfrac%7BmM%7D%7Br%7D%3D-%286.67%5Ccdot%2010%5E%7B-11%7D%29%5Cfrac%7B%28525%20kg%29%285.98%5Ccdot%2010%5E%7B24%7Dkg%29%7D%7B6.37%5Ccdot%2010%5E6%20m%7D%3D-3.29%5Ccdot%2010%5E%7B10%7D%20J)
And since it hits the ground with speed
![v=1.90 km/s = 1900 m/s](https://tex.z-dn.net/?f=v%3D1.90%20km%2Fs%20%3D%201900%20m%2Fs)
it also has kinetic energy:
![K=\frac{1}{2}mv^2=\frac{1}{2}(525 kg)(1900 m/s)^2=9.48\cdot 10^8 J](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2%3D%5Cfrac%7B1%7D%7B2%7D%28525%20kg%29%281900%20m%2Fs%29%5E2%3D9.48%5Ccdot%2010%5E8%20J)
So the total energy when the satellite hits the ground is
![E_f = U+K=-3.29\cdot 10^{10}J+9.48\cdot 10^8 J=-3.20\cdot 10^{10} J](https://tex.z-dn.net/?f=E_f%20%3D%20U%2BK%3D-3.29%5Ccdot%2010%5E%7B10%7DJ%2B9.48%5Ccdot%2010%5E8%20J%3D-3.20%5Ccdot%2010%5E%7B10%7D%20J)
So the energy transformed into internal energy due to air friction is the difference between the total initial energy and the total final energy of the satellite:
![\Delta E=E_i-E_f=-1.51\cdot 10^{10} J-(-3.20\cdot 10^{10} J)=1.69\cdot 10^{10}J](https://tex.z-dn.net/?f=%5CDelta%20E%3DE_i-E_f%3D-1.51%5Ccdot%2010%5E%7B10%7D%20J-%28-3.20%5Ccdot%2010%5E%7B10%7D%20J%29%3D1.69%5Ccdot%2010%5E%7B10%7DJ)