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slamgirl [31]
4 years ago
13

In the SI unit system, time (t) is measured in seconds, distance (d) is measured in meters, velocity (v) is measured in meters p

er second, and acceleration (a) is measured in meters per second squared.
Based on these units, select any of the following expressions that could be valid expressions?
1- t=cubed rootinside d^2/va
2- v=at
3- a=sqrt vd/t^3
4- d=at
5- v=a/t
6- a=d/t^2
Mathematics
1 answer:
LenaWriter [7]4 years ago
5 0

Answer:

1) t=cubed rootinside d^2/va

2) v=at              

3) a=sqrt vd/t^3  

6) a=d/t^2                                      

Step-by-step explanation:

We are given the following:

In the SI unit system, time (t) is measured in seconds(s),

distance (d) is measured in meters(m)

velocity (v) is measured in meters per second(ms^{-1}), and

acceleration (a) is measured in meters per second squared(ms^{-2}).

\text{Velocity} = \dfrac{\text{Distance}}{\text{Time}}\\\\\text{Acceleration} = \displaystyle\frac{\text{Change in velocity}}{\text{Time}}\\\\1)\\\\t = (\dfrac{d^2}{va})^{\frac{1}{3}}\\\\t = (\dfrac{m^2}{ms^{-1}ms^{-2}})^{\frac{1}{3}}\\\\t = (s^3)^{\frac{1}{3}}\\\\t = s\\2)\\v = at\\v = ms^{-2}\times s\\v = ms^{-1}\\\\3)\\a = \sqrt{\frac{vd}{t^3}}\\\\a = \sqrt{\frac{(ms^{-1})(m)}{s^3}}\\\\a = \sqrt{m^2s^{-4}} = ms^{-2}\\\\6)\\a = \frac{d}{t^2}\\\\a = \frac{m}{s^2} = ms^{-2}

Thus, valid expressions are

1) t=cubed rootinside d^2/va

2) v=at              

3) a=sqrt vd/t^3  

6) a=d/t^2      

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