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Verdich [7]
2 years ago
15

2. A car accelerates uniformly from +10.0 m/s to +50.0 m/s over a distance of 225 m. How long did it take to go that distance? S

how all your work, including the equation used, given and unknown quantities, and any algebra required. Make sure your answer has the correct number of significant figures.
Physics
1 answer:
artcher [175]2 years ago
7 0
Let's call the constant acceleration a.
At a time t, its speed will thus be v(t)=a*t+v0 where v0 is its initial speed, here 10 m/s. Hence v(t)=a*t+10.

From there we can deduce the position P(t)=a*t^2/2+10t+p0 where p0 is the initial position, here 0.

Hence P(t)=a*t^2/2+10t

Let's call T the time at which it's at 50 m/s, we know that P(T)=225m and that v(T)=50 m/s hence a*T+10=50 thus a=40/T and P(T)=(40/2+10)T=30T

Hence T=225/30=7.5

It took 7.5 seconds


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zubka84 [21]

Answer:

Therefore, the revolutions that each tire makes is:

\Delta \theta=22\: rev

Explanation:

We can use the following equation:

\omega_{f}^{2}=\omega_{i}^{2}-2\alpha \Delta \theta (1)

The angular acceleration is:

a_{tan}=\alpha R

\alpha=\frac{1.9}{0.325}

\alpha=5.85\: rad/s^{2}

and the initial angular velocity is:

\omega_{i}=\frac{v}{R}

\omega_{i}=\frac{27.2}{0.325}

\omega_{i}=83.69\: rad/s

Now, using equation (1) we can find the revolutions of the tire.

0=83.69^{2}-2*25.85 \Delta \theta

\Delta \theta=135.47\: rad

Therefore, the revolutions that each tire makes is:

\Delta \theta=22\: rev

I hope it helps you!

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3 years ago
What is angle of dip​
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Are photons causing the electrons to flow off of the target? ( the target is calcium)
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A ball hits a wall horizontally at 6m/s and rebounces at 4.4m/s the ball is in contact with wall for 0.04 sec. what is the accel
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Answer:

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