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ikadub [295]
3 years ago
8

A car traveling at 15 m/s starts to decelerate steadily. It comes to a complete stop in 10 seconds. What is it’s acceleration

Physics
2 answers:
tresset_1 [31]3 years ago
7 0

So the initial velocity is 15 m/s, the final velocity is 0 since it's at a complete stop and time is 10 seconds.  Therefore:

Acceleration=\frac{v_{final}-v_{initial}}{t_{final}-t_{initial}} =\frac{0m/s-15m/s}{10s-0s} =-1.5\frac{m}{s^2}

Therefore, the acceleration is -1.5 m/s^2.  The reason it's negative is due to the fact that the vector is going against it's original movement since it's decelerating.

____ [38]3 years ago
3 0

Hello!

A car traveling at 15 m/s starts to decelerate steadily. It comes to a complete stop in 10 seconds. What is it's acceleration ?

We have the following data:

Vi (initial velocity) = 15 m/s (starts)

Vf (final velocity) = 0 m/s (stop)

t (time) = 10 s

a (acceleration) = ? (in m/s²)

<u><em>We apply the data to the formula of the hourly function of the velocity, let us see:</em></u>

V_f = V_i + a*t

0 = 15 + a*10

- 15 = 10\:a

10\:a = - 15

a = \dfrac{-15}{10}

\boxed{\boxed{a = - 1.5\:m/s^2}}\Longrightarrow(the\:car\:slows\:down)\:\:\:\:\:\:\bf\green{\checkmark}

<u><em>Answer:  </em></u>

<u><em>The acceleration is -1.5 m/s² (decelerate)  </em></u>

________________________________

\bf\red{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}

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Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The magnitude is   B= 4.2 *10^ {-6}T , the direction is into the page

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         B_{da} = \frac{\mu_o I}{4 \pi r_{da}} \theta

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                   B_{da}= 12.56*10^{-6} T

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       Since B_{da} > B_{bc} then the direction of the net charge would be into the page

 

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