Answer:
The average speed of the earth in its orbit is ![29.86km/s](https://tex.z-dn.net/?f=29.86km%2Fs)
Explanation:
The average distance between the Earth and the Sun is
.
The average speed of the earth in its orbit can be found by the next equation :
(1)
Where r is the radius and T is the period.
In this case, the orbit of the Earth can be considered as a circle
(
) instead of an ellipse.
It takes 1 year to the Earth to make one revolution around the Sun. Therefore, its period will be 365.25 days.
Notice that to express the period in terms of seconds, the following is needed:
⇒ ![31557600s](https://tex.z-dn.net/?f=31557600s)
Then, equation 1 can be used:
![v = \frac{2 \pi (1.50x10^{8}km)}{31557600s}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B2%20%5Cpi%20%281.50x10%5E%7B8%7Dkm%29%7D%7B31557600s%7D)
![v = 29.86km/s](https://tex.z-dn.net/?f=v%20%3D%2029.86km%2Fs)
The
sun is a ball of hot gases containing different kinds of elements at different
cores. It has a very high temperature that radiates all throughout the Milky
Way galaxy. The sun has three main parts; photosphere, chromospheres
and corona. The outer core of a star located at the chromospheres contains
mostly of hydrogen. Inside the hydrogen is helium then carbon, oxygen, neon,
magnesium silicon and the inert gas. The photosphere is scattered by the loose electrons in the corona’s plasma.
Answer:
The net change in the internal energy of the gas in the piston is -343J
Explanation:
Because heat and workdone are the only means of energy transfer between the system and the surrounding, change in internal energy is given by;
∆E = q + w
q = heat transfer
w = workdone
Because heat is lost by the system, the heat transfer is negative
q = -413J
Because work is done on the system, workdone is positive
w = +70J
∆E = -413J + 70J
∆E = -343J
Answer:
The dog ran a total distance of 45m but he is only 5m away from the starting line
Explanation: When you add 25 to 20 you get 45 for the total distance and if he ran back in the same direction then you would subtract 20 from 25 and get 5m
Answer:
Total pressure exerted at bottom = 119785.71 N/m^2
Explanation:
given data:
volume of water in bottle = 150 L = 0.35 m^3
Area of bottle = 2 ft^2
density of water = 1000 kg/m
Absolute pressure on bottom of bottle will be sum of atmospheric pressure and pressure due to water
Pressure due to water P = F/A
F, force exerted by water = mg
m, mass of water = density * volume
= 1000*0.350 = 350 kg
F = 350*9.8 = 3430 N
A = 2 ft^2 = 0.1858 m^2
so, pressure P = 3430/ 0.1858 = 18460.71 N/m^2
Atmospheric pressure
At sea level atmospheric pressure is 101325 Pa
Total pressure exerted at bottom = 18460.71 + 101325 = 119785.71 N/m^2
Total pressure exerted at bottom = 119785.71 N/m^2