Answer:
Blue supergiants represent a slower burning phase in the death of a massive star. Due to core nuclear reactions being slightly slower, the star contracts and since very similar energy is coming from a much smaller area (photosphere) then the star's surface becomes much hotter.
Explanation:
I know this may not be the answer youre looking for, but hopefully this can help somehow!
The material that the snails are moving on affects their speed. On a smooth material, (petri dish) the snails moved the fastest. On a rough material (sandpaper) the snails moved the slowest. On dirt, a compromise between smooth and rough, the snails moved at a medium pace. The material and possibly friction affect this.
If a football is kicked from the ground with a speed of 16.71 m/s at an angle of 49.21 degrees, then the vertical component of the initial velocity would be 12.65 m/s
<h3>What is Velocity?</h3>
The total displacement covered by any object per unit of time is known as velocity. It depends on the magnitude as well as the direction of the moving object. The unit of velocity is meter/second.
As given in the problem A football is kicked from the ground with a speed of 16.71 m/s at an angle of 49.21 degrees
The horizontal component of the velocity is given by
Vx = Vcosθ
The vertical component of the velocity is given by
Vy = Vsinθ
As we have to find the vertical component of the velocity given that speed of 16.71 m/s at an angle of 49.21 degrees from the ground
Vy = 16.71 × sin49.21°
Vy = 12.65 m/s
Thus, the vertical component of the velocity would be 12.65 m/s
Learn more about Velocity from here
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Answer:
C) one-half as great
Explanation:
We can calculate the acceleration of gravity in that planet, using the following kinematic equation:
In this case, the sphere starts from rest, so 
. Replacing the given values and solving for g':
The acceleration due to gravity near Earth's surface is 
. So, the acceleration due to gravity near the surface of the planet is approximately one-half of the acceleration due to gravity near Earth's surface.
