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alexdok [17]
3 years ago
8

A graduated cylinder contains 63.0 mL of water. A piece of gold, which has a density of 19.3 g/ cm3, is added to the water and t

he volume goes up to 64.5 mL. Calculate the mass in grams of the gold that was added to the water. Explain how you got your answer.
Physics
1 answer:
icang [17]3 years ago
7 0

Answer:

29.0 g

Explanation:

The mass of the piece of gold is given by:

m = dV

where

m is the mass

d is the density

V is the volume of the piece of gold

The density of gold is

d = 19.3 g/cm^3

while the volume of the sample is equal to the volume of displaced water, so

V = 64.5 mL - 63.0 mL = 1.5 mL

And since

1 mL = 1 cm^3

the volume is

V = 1.5 cm^3

So the mass of the piece of gold is:

m = (19.3 g/cm^3)(1.5 cm^3)=29.0 g

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A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The
Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}

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\sum {F = ma}

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

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F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg

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Use the expression of static friction and solve for maximum static friction for box of mass {m_1}

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{F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute  for , [/tex]{m_1}[/tex] for in the equation .

{F_{\rm{f}}} = {m_1}a

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{\mu _{\rm{s}}}{m_1}g = {m_1}a

Rearrange for a.

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The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

{F_{\min }} = \left( {{m_1} + {m_2}} \right)a

Substitute for in the above equation .

{F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

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Answer: A

<u>Explanation:</u>

NOTES:

d = 650 meters

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**********************************

v = d/t

  = 650 meters/10 seconds

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