Answer:
The modern instruments or we can say the different levels of telescopes are used to explore and study the distant galaxies. i.e the Hubble telescope is out there providing the data regarding the different properties of the celestial entities which in other case is not visible to the human naked eye.
Explanation:
- Scientists and research workers are in constant search for more answers as they explore the universe and implement the laws of physics on the celestial entities. But, most of the objects inside the universe are not visible to human naked eye, as they are far from sight and thus more advanced form of instruments like the x-ray, optical, and light telescopes are used to determine the different properties of the celestial entities inside the universe.
- As, these telescopes includes the most recent "Hubble telescope", which is out there inside the space to explore the universe and more over the galaxies by subjecting them with x-rays and then provide us with a very rough but valid results to study the distant galaxies.
Explanation:
The momentum of the three objects are as follow :
11 kg-m/s, -65 kg-m/s and -100 kg-m/s
Before collision, the momentum of the system is :
After collison, they move together. It means it is a case of inelastic collision. In this type of collision, the momentum of the system remains conserved.
It would mean that, after collision, momentum of the system is equal to the initial momentum.
Hence, final momentum = -154 kg-m/s.
At the equator, the North Celestial and South Celestial Poles would lie on the horizon where the meridian intersects the horizon. Polaris is called the "North Star." ... Polaris is special because the Earth's North Pole points almost exactly towards it.
The conversion from gallons to liters is 1 = 3.785.
Keeping this in mind...
42 x 3.785 = 158.97 liters.
If rounding, there are about 159 liters of oil in a barrel.
Answer:
I_v = 2,700 W / m^2
I_m = 610 W / m^2
I_s = 16 W / m^2
Explanation:
Given:
- The Power of EM waves emitted by Sun P_s = 4.0*10^26 W
- Radius of Venus r_v = 1.08 * 10^11 m
- Radius of Mars r_m = 2.28 * 10^11 m
- Radius of Saturn r_s = 1.43 * 10^12 m
Find:
Determine the intensity of electromagnetic waves from the sun just outside the atmospheres of (a) Venus, (b) Mars, and (c) Saturn.
Solution:
- We know that Power is related to intensity and surface area of an object follows:
I = P / 4*pi*r^2
Where, A is the surface area of a sphere models the atmosphere around the planets.
a)
- The intensity at the surface of Venus is calculated as:
I_v = P_s / 4*pi*r^2_v
I_v = 4.0*10^26 / 4*pi*(1.08*10^11)^2
I_v = 2,700 W / m^2
b)
- The intensity at the surface of Mars is calculated as:
I_m = P_s / 4*pi*r^2_m
I_m = 4.0*10^26 / 4*pi*(2.28*10^11)^2
I_m = 610 W / m^2
c)
- The intensity at the surface of Saturn is calculated as:
I_s = P_s / 4*pi*r^2_s
I_s = 4.0*10^26 / 4*pi*(1.43*10^12)^2
I_s = 16 W / m^2
