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1 year ago

PLEASE HELP!!!!

Physics

2 answers:

Goryan [66]1 year ago

7 0

Answer:

The braking distance would be about nine times as long (assuming that acceleration during braking stays the same.)

Explanation:

Let $u$ denote the initial velocity of the vehicle ( $20\; \text{mph}$ or $60\; \text{mph}$ ) and let $v$ denote the velocity of the vehicle after braking ( $0\; \text{mph}$ ). Let $x$ denote the braking distance.

Assume that the acceleration during braking are both constantly $a$ in both scenarios. The SUVAT equations would apply. In particular:

$\begin{aligned} x &= \frac{v^{2} - u^{2}}{2\, a}\end{aligned}$ .

Since $v = 0$ <em> </em>(the vehicle has completely stopped), the equation becomes $x = (-u^{2}) / (2\, a)$ .

Assuming that $a$ (braking acceleration) stays the same, the braking distance $x$ would be proportional to $u^{2}$ , the square of the initial velocity.

Hence, increasing the initial speed from $20\; \text{mph}$ to $60\; \text{mph}$ would increase the braking distance by a factor of $3^{2} = 9$ .

netineya [11]1 year ago

7 0

Answer:

9 times

Explanation:

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Answer:

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Answer:

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Explanation:

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Finding the ratio between the two brightness we get:

$\displaystyle{\frac{B_{Betelgeuse}}{B_{Sirius}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (26\ ly)^2}{26L_{Sun}} \approx 1.43 \cdot 10^{6} }$

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