Answer:
Explanation:
what is the smallest crater that each of these telescopes could resolve on our moon?
For moon ;
s = 3.8 × 10 ⁸ m
y = 1.22 λs/D
where;
λ = 400 nm = 400× 10 ⁻⁹
D = 2.4 m
The smallest crater for the hubble space is calculated as follows:
For Aceribo ;
y = 1.22 λs/D
where :
λ = 75 cm = 0.75 m
D = 305 m
Answer:
option (b) 4900 N
Explanation:
m = 2000 kg, R = 6380 km = 6380 x 10^3 m, Me = 5.98 x 10^24 kg, h = R
F = G Me x m / (R + h)^2
F = G Me x m / 2R^2
F = 6.67 x 10^-11 x 5.98 x 10^24 x 2000 / (2 x 6380 x 10^3)^2
F = 4900 N
Explanation:
When you observe the night sky you will notice that the stars are moving. They rise from eastern horizon and set in the western horizon. It happens due to rotation of Earth. When observed closely you will notice that the all the stars seem to go around the pole star. Out of all the stars there are some stars which neither set not rise, such stars are called as Circumpolar stars. This means that they are always above the horizon. If we trace the path of such stars they will appear to make complete circle around the pole star.
Also, you will notice that the altitude of pole star (separation of pole star from the horizon in degrees) will depend on the location of observe on the Earth. This happens due to Earth being spherical. So if you are on equator the pole star will be on the horizon i.e. 0° altitude. If you are at Poles, altitude of the pole star will be 90°. Technically the altitude of pole star at any place on Earth is equal to the latitude of the place.
If the altitude of pole star varies and increases as you move towards higher latitude on Earth, the distance between horizon and pole star will also increase. This will result in more stars being circumpolar.
If you are at Poles, all the stars will be circumpolar and if you are at equator no star will be circumpolar.
Answer:
Explanation:
<u>Uniform Acceleration
</u>
When an object changes its velocity at the same rate, the acceleration is constant.
The relation between the initial and final speeds is:
Where:
vf = Final speed
vo = Initial speed
a = Constant acceleration
t = Elapsed time
It's known a train moves from rest (vo=0) to a speed of vf=25 m/s in t=30 seconds. It's required to calculate the acceleration.
Solving for a:
Substituting:
