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3 years ago

A cheetah can accelerate from rest to a speed of 21.5 m/s in 6.75 s. What is its acceleration? m/s^2

Physics

2 answers:

frez [133]3 years ago

8 0

Answer:

$a=3.185\frac{m}{s^2}$

Explanation:

Acceleration is the change in velocity for a given period of time, we can express this in the next formula:

$a = \frac{\Delta v}{\Delta t} =\frac{v_{1}-v_{0}}{t_{1}-t_{0}}$

In this case the values are:

$v_{0}=0\\v_{1}= 21.5 m/s\\t_{0}=0\\t_{1}= 6.75 s\\$

Inserting known values, the acceleration is:

$a= \frac{21.5 m/s}{6.75 s} \\a=3.185\frac{m}{s^2}$

grandymaker [24]3 years ago

6 0

Answer:

Acceleration will be $a=3.185m/sec^2$

Explanation:

We have given final velocity v = 21.5 m/sec

Time t = 6.75 sec

As cheetah starts from rest so initial velocity u = 0 m/sec

From first equation of motion we know that v = u+at, here v is final velocity, u is initial velocity, a is acceleration and t is time

So $21.5=0+a\times 6.75$

$a=3.185m/sec^2$

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Explanation:

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To make the numbers easier to work with, round each atomic weight. We'll say the atomic weight of potassium is 39 and the atomic weight of argon is 40. To see how many neutrons each one has, I can set up a simple equation for each using the following equation:

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