Answer:
a cold air mass and a warm air mass merge together
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:
<h2>The pin's final velocity is 5m/s</h2>
Explanation:
Step one:
given data
mass of ball m1=5kg
initial velocity of ball u1=10m/s
mass of pin m2=2kg
initial velocity of pin u2= 0m/s
final velocity of ball v2=8m/s
final velocity of pin v2=?
Step two:
The expression for elastic collision is given as
m1u1+m2u2=m1v1+m2v2
substituting we have
5*10+2*0=5*8+2*v2
50+0=40+2v2
50-40=2v2
10=2v2
divide both sides by 2
v2=10/2
v2=5m/s
The pin's final velocity is 5m/s
Answer:
b. 14
Explanation:
= Initial temperature = 27 °C = 27 + 273 = 300 K
= Final temperature = 37 °C = 37 + 273 = 310 K
= Initial Power radiated by the object
= Final Power radiated by the object
We know that the power radiated is directly proportional to fourth power of the temperature. hence
Percentage increase in power is given as
