an athlete runs 5.4 laps around a circular track that is 400.0 m long. If this takes 540 s, what is the average velocity of the
1 answer:
Answer:
Average velocity is 0.296 m/s.
Average speed is 4.0 m/s.
Explanation:
Given:
Distance of the circular track is,
Number of laps ran is,
Time taken for the run is,
Now, total distance covered in 5.4 laps =
Also, since the path is a circle, the final position of the athlete after 5.4 laps will be 0.4 of 400 m ahead of the starting point.
Distance covered in 0.4 laps is,
Therefore, the displacement of the athlete will be 160 m as the athlete is 160 m ahead of the starting point and displacement depends on the initial and final points only.
Now, average velocity is given as:
Average speed is the ratio of total distance covered to total time taken.
So, average speed =
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Answer:
2.43J
Explanation:
Given parameters:
Mass of the arrow = 0.155kg
Velocity = 31.4m /s
Unknown:
Kinetic energy when it leaves the bow = ?
Solution:
The kinetic energy of a body is the energy in motion of the body;
it can be derived using the expression below:
K.E = m v²
m is the mass
v is the velocity
Solve for K.E;
K.E = x 0.155 x 31.4 = 2.43J
Answer:
b) Betelgeuse would be times brighter than Sirius
c) Since Betelgeuse brightness from Earth compared to the Sun is the statement saying that it would be like a second Sun is incorrect
Explanation:
The start brightness is related to it luminosity thought the following equation:
(1)
where is the brightness,
is the star luminosity and
, the distance from the star to the point where the brightness is calculated (measured). Thus:
b) and
where
is the Sun luminosity (
) but we don't need to know this value for solving the problem.
is light years.
Finding the ratio between the two brightness we get:
c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is . Then
Notice that since the star luminosities are given with respect to the Sun luminosity we don't need to use any value a simple states the Sun luminosity as the unit, i.e 1. From this result, it is clear that when Betelgeuse explodes it won't be like having a second Sun, it brightness will be 5 orders of magnitude smaller that our Sun brightness.