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Ksju [112]
3 years ago
12

A package of aluminum foil contains 50. ft2 of foil, which weighs approximately 6.0 oz. Aluminum has a density of 2.70 g/cm3. Wh

at is the approximate thickness of the foil in millimeters? Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
SIZIF [17.4K]3 years ago
3 0

Answer:

1.36 x 10^-3 cm

Explanation:

Area = 50 ft^2 = 46451.5 cm^2

mass = 6 oz = 170.097 g

density = 2.70 g/cm^3

Let t be the thickness of foil in cm.

mass = volume x density

mass = area x thickness x density

170.097 = 46451.5 x t x 2.70

t = 1.36 x 10^-3 cm

Thus, the thickness of aluminium foil is 1.36 x 10^-3 cm.

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A flask that weighs 345.8 g is filled with 225 mL of carbon tetrachloride.
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Answer:

1.59 g/mL

Explanation:

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m = 703.55 g − 345.8 g

m = 357.75 g

Density = mass / volume

d = 357.75 g / 225 mL

d = 1.59 g/mL

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Ryan is experimenting with core materials for an electromagnet. He slides different core materials through a coil of current-car
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When is your kinetic energy the least when swinging on a park swing?
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Let's say you are on the third story of an apartment building that has a swimming pool. For some insane reason you think it is a
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\displaystyle v_o=x.\sqrt{\frac{g}{2y}}

Explanation:

Horizontal Launch

When an object is thrown with a specified initial speed in the horizontal direction, it describes a curved path that finishes when it hits the ground level after traveling certain horizontal distance x and a vertical height y from the launching point. The horizontal speed is always constant and the vertical speed increases due to the effect of gravity. It can be found that the horizontal distance reached by the object when launched at an initial speed  in a given time t is

x=v_o.t

And the vertical distance is

\displaystyle y=\frac{g.t^2}{2}

If t is the total flight time, then x and y are maximum and we can find a relation between them. Solving for t in the first equation

\displaystyle t=\frac{x}{v_o}

Substituting in the second equation

\displaystyle y=\frac{g}{2}\left ( \frac{x}{v_o}\right )^2

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\displaystyle \left ( \frac{v_o}{x}\right )^2=\frac{g}{2y}

Solving for v_o

\displaystyle v_o=x.\sqrt{\frac{g}{2y}}

There are many applications for the horizontal launch. One common situation is when someone wants to drop something on certain terrain at a specific approximate point when traveling in a plane at a given height. Once the object is left fall, it has the same speed as the plane, so the plane speed can be estimated to make the best possible launch, or given that speed, we can know in advance where the object will reach ground level

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