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nikitadnepr [17]
4 years ago
7

A race car circles 10 times around a circular 8.0-km track in 20 min.

Physics
2 answers:
Otrada [13]4 years ago
6 0
Find the average speed and the average velocity.

Average speed = distance / time

distance = 10 x 8000 m = 80,000 m
time = 20 min * 60 s/min = 1200 s

Average speed = 80,000 m / 1200 s = 66.67 m/s

Average velocity = displacement / time

Given that the race car made complete circles the final poin is the same initial point, then its displacement is zero and the average velocity is zero too.
fredd [130]4 years ago
5 0
Assuming that you're asking for the average speed, 

Average speed  = distance / time

since he circle 8 km track 10 times, the distance is 80,000 m and the time is 20 min or 1200 seconds
Average speed = 80,000 / 1200   = 66.67 m/s

Hope this helps
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The acorn was at a height of <u>4.15 m</u> from the ground before it drops.

The acorn takes a time t to fall through a distance h₁, which is the length of the scale. When the acorn reaches the top of the scale, its velocity is u.

Calculate the speed of the acorn at the top of the scale, using the equation of motion,

s=ut+ \frac{1}{2} at^2

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s=ut+ \frac{1}{2} at^2\\ (2.27 m)=u(0.301s)+\frac{1}{2}(9.8m/s^2)(0.301s)^2\\ u=\frac{1.8261m}{0.301s} =6.067m/s

If the acorn starts from rest and reaches a speed of 6.067 m/s at the top of the scale, it would have fallen a distance h₂ to achieve this speed.

Use the equation of motion,

v^2=u^2+2as

Substitute 6.067 m/s for v, 0 m/s for u, 9.8 m/s² for a (=g) and h₂ for s.

v^2=u^2+2as\\ (6.067m/s)^2=(0m/s)^2+2(9.8m/s^2)h_2\\ h_2=\frac{(6.067m/s)^2}{2(9.8m/s^2)} =1.878 m

The height h above the ground at which the acorn was is given by,

h=h_1+h_2=(2.27 m)+(1.878 m)=4.148 m

The acorn was at a height <u>4.15m</u> from the ground before dropping down.

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