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2 years ago

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Superman is standing 393 m horizontally away from Lois Lane. A villain drops a rock from 4.00 m directly above Lois. If Superman

is to intervene and catch the rock just before it hits Lois, what should be his minimum constant acceleration? If the acceleration is toward Lois, enter a positive value. If the acceleration is away from Lois, enter a negative value.

1 answer:

Sergio039 [100]2 years ago

6 0

Answer:

-963.93 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 4=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{4\times 2}{9.81}}\\\Rightarrow t=0.903\ s$

$s=ut+\frac{1}{2}at^2\\\Rightarrow 393=0\times 0.0903+\frac{1}{2}\times a\times 0.903^2\\\Rightarrow a=\frac{393\times 2}{0.903^2}\\\Rightarrow a=963.93\ m/s^2$

The acceleration of Superman would be -963.93 m/s² from Lois' perspective

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Answer:

a) the mug hits the floor 0.7425m away from the end of the bar. b) |V|=5.08m/s θ= -72.82°

Explanation:

In order to solve this problem, we must first start by doing a drawing of the situation. (see attached picture).

a)

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In order to find the time we need to use the following formula, which contains the data we know:

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we know that $y_{f}=0$ and that $v_{y0}=0$ as well, so the formula is simplified to:

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we can now solve this for t, so we get:

$-y_{0}=\frac{1}{2}at^{2}$

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which yields:

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so we can solve this for x, so we get:

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and we can now substitute the values we know:

x=(1.5m/s)(0.495s)

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so we can solve this for the final velocity:

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$V_{yf}=(-9.8m/s^{2})(0.495s)$

which yields:

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so now we know the components of the final velocity, which are:

$V_{xf}=1.5m/s$ and $V_{yf]=-4.851m/s$

so now we can find the speed by determining the magnitude of the vector, like this:

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so we get:

$|V|=\sqrt{(1.5m/s)^{2}+(-4.851m/s)^{2}$

which yields:

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$\theta = tan^{-1} (\frac{V_{y}}{V_{x}})$

so we get:

$\theta = tan^{-1} (\frac{-4.851m/s}{(1.5m/s)})$

which yields:

$\theta = -72.82^{o}$

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