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2 years ago

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An empty bottle has a mass of 35.00 grams. When filled with water, it has a mass of 98.44 grams. Of the same bottle is filled wi

th a different fluid, the mass is 89.22 grams. What is the specific gravity of this other fluid

1 answer:

sveticcg [70]2 years ago

5 0

Answer:

Specific gravity of other fluid = .854 (Approx)

Explanation:

Given:

Mass of water = 35 g

Mass of filled bottle with water = 98.44 g

Mass of filled bottle with fluid = 89.22 g

Computation:

Mass of water = 98.44g - 35g = 63.44g

Density of water = 1000 g/L

Volume of bottle = 63.44/1000 = 0.06344L

Mass of other liquid = 89.22g - 35g = 54.22g

Density of other liquid = 54.22g/0.06344L = 854.665826 g/L

Water has a specific gravity = 1

So , specific gravity of other fluid

1000 / 854.665826 = 1 / specific gravity of other fluid

Specific gravity of other fluid = .854 (Approx)

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Answer:

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Explanation:

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

a)

From the drawing we can see that we are dealing with a two dimensions movement problem. So in order to find out how far away from the bar the mug will fall, we need to start by finding how long it will take the mug to be in the air, so we analyze the vertical movement of the mug.

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:

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

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

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

$-2y_{0}=at^{2}$

$\frac{-2y_{0}}{a}=t^{2}$

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$t=\sqrt{\frac{-2(1.20m)}{(-9.8m/s^{2})}}$

which yields:

t=0.495s

we can now use this to find the horizontal distance the mug travels. We know that:

$V_{x}=\frac{x}{t}$

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

$x=V_{x}t$

and we can now substitute the values we know:

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

which yields:

x=0.7425m

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$a=\frac{v_{f}-v_{0}}{t}$

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$a=\frac{v_{f}}{t}$

so we can solve this for the final velocity:

$V_{yf}=at$

in this case the acceleration is the same as the acceleration of gravity (which is negative) so we can substitute that and the time we found on the previous part to get:

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

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

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

which yields:

|V|=5.08m/s

now, to find the direction of the impact, we can use the following equation:

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

so we get:

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

which yields:

$\theta = -72.82^{o}$

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