Answer:
v = 3.08 km/s
Explanation:
Given that,
The angular velocity of the satellite = 
A satellite is in orbit 36000km above the surface of the earth.
The radius of the earth is 6400 km
Let v is the velocity of the satellite. It can be calculated as :
So, the velocity of the satellite is 3.08 km/s.
Answer:
E- The star becomes a red giant (LATEST STAGE)
F- The surface of the star becomes brighter and cooler
C- Pressure from the star's hydrogen-burning shell causes the non burning envelope to expand
A- The shell of hydrogen surrounding the star's nonburning helium core ignites.
D- The star's non burning helium core starts to contract and heat up
B- Pressure in the star's core decreases (EARLIEST STAGE)
(A star moves away from the main sequence once its core runs out of hydrogen to fuse into helium. The energy once supplied by hydrogen burning reduces and the core starts to compress under the force of gravity. This contraction allows the core and surrounding layers to heat up. Finally, the hydrogen shell around the core becomes hot enough to ignite hydrogen burning.
For the answer to the question above, first find out the gradient.
<span>m = rise/run </span>
<span>=(y2-y1)/(x2-x1) </span>
<span>the x's and y's are the points given: "After three hours, the velocity of the car is 53 km/h. After six hours, the velocity of the car is 62 km/h" </span>
<span>(x1,y1) = (3,53) </span>
<span>(x2,y2) = (6,62) </span>
<span>sub values back into the equation </span>
<span>m = (62-53)/(6-3) </span>
<span>m = 9/3 </span>
<span>m = 3 </span>
<span>now we use a point-slope form to find the the standard form </span>
<span>y-y1 = m(x-x1) </span>
<span>where x1 and y1 are any set of point given </span>
<span>y-53 = 3(x-3) </span>
<span>y-53 = 3x - 9 </span>
<span>y = 3x - 9 + 53 </span>
<span>y = 3x + 44 </span>
<span>y is the velocity of the car, x is the time.
</span>I hope this helps.
Answer:
Haven't done this but I think it will increase by 73.29% not too anyone correct me if I'm worng
<span>U could compare them using the intensity
technique when bending waves are negligible in comparison with
quasi-longitudinal waves.</span>
