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Gold

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Subject

Physics

School

Florida Institute of Technology Melbourne

Type

Homework

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Week 8 Homework

Fluid Statics

Name: Grade:

Must show all work for full credit

Answers must include units if required and highlighted or in bold

(if these rules are not met, they will be considered wrong)

1) Gold is sold by the troy ounce (31.103 g). What is the volume of 1 troy ounce of pure

gold?

Ans:

Given,

Mass of 1 troy of gold(M) = 31.103 g

Density of pure gold (𝜌) = 19.3 g/cc

Therefore,

Volume of 1 troy of pure gold = M/ 𝜌

= 31.103/19.3

= 1.6115 cc

2) Mercury is commonly supplied in flasks containing 34.5 kg (about 76 lb). What is the

volume in liters of this much mercury?

Ans:

Given,

Mass of mercury contained in the flask (M) = 34.5 kg

Density of mercury (𝜌) = 13600 kg/m

Therefore,

Volume of mercury in the flask = M/ 𝜌

= 34.5/13600

= 2.5367*10

-3

= 2.5367 liters

3) What is the mass of a deep breath of air having a volume of 2.00 L? (b) Discuss the effect

taking such a breath has on your body's volume and density.

Ans:

Given,

Volume of air (V) = 2 L = 0.002 m

Density of air (𝜌) = 1.225 kg/m

Therefore,

Mass of a deep breath of air = V* 𝜌

= 0.002*1.225

= 2.45*10

-3

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Week 8 Homework

Fluid Statics

4) A straightforward method of finding the density of an object is to measure its mass and

then measure its volume by submerging it in a graduated cylinder. What is the density of

a 240-g rock that displaces 29.0𝑐𝑚

of water? (Note that the accuracy and practical

applications of this technique are more limited than a variety of others that are based on

Archimedes’ principle.)

Ans:

Given,

Mass of the rock (M) = 240 g = 0.24 kg

Volume of rock (V) = 29 cm

= 2.9*10

-5

Therefore,

Density of the rock (𝜌) = M/V

= 0.24/2.9*10

-5

= 8275.8620 kg/m

5) (a) A rectangular gasoline tank can hold 50.0 kg of gasoline when full. What is the depth

of the tank if it is 0.500-m wide by 0.900-m long? (b) Discuss whether this gas tank has a

reasonable volume for a passenger car.

Ans:

Given,

Mass of gas that can be hold by gasoline tank(M) = 50kg

Density of gasoline (𝜌) = 800 kg/m

(a)

Therefore,

Volume of gasoline tank (V) = M/ 𝜌

= 50/800

= 0.0625 m

or 62.5 liters

Then, let d be depth of the tank which is 0.500-m wide and 0.900-m long.

Also,

Volume of the tank (V) = width*length*depth

Or, 0.0625 = 0.5*0.9*d

So, d = 0.13889 m

Therefore, depth of the tank is 0.13889 m.

(b) Since, volume of gas tank of a passenger car must be between 45 and 65 liters and the tank

we have has a capacity of 62.5 liters. So, the tank we have has a reasonable volume for a

passenger car.

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Week 8 Homework Fluid Statics Name: Grade: Must show all work for full credit Answers must include units if required and highlighted or in bold (if these rules are not met, they will be considered wrong) 1) Gold is sold by the troy ounce (31.103 g). What is the volume of 1 troy ounce of pure gold? Ans: Given, Mass of 1 troy of gold(M) = 31.103 g Density of pure gold (𝜌) = 19.3 g/cc Therefore, Volume of 1 troy of pure gold = M/ 𝜌 = 31.103/19.3 = 1.6115 cc 2) Mercury is commonly supplied in flasks containing 34.5 kg (about 76 lb). What is the volume in liters of this much mercury? Ans: Given, Mass of mercury contained in the flask (M) = 34.5 kg Density of mercury (𝜌) = 13600 kg/m3 Therefore, Volume of mercury in the flask = M/ 𝜌 = 34.5/13600 = 2.5367*10-3 m3 = 2.5367 liters 3) What is the mass of a deep breath of air having a volume of 2.00 L? (b) Discuss the effect taking such a breath has on your body's volume and density. Ans: Given, Volume of air (V) = 2 L = 0.002 m3 Density of air (𝜌) = 1.225 kg/m3 Therefore, Mass of a deep breath of air = V* 𝜌 = 0.002*1.225 = 2.45*10-3 kg Week 8 Homework Fluid Statics 4) A straightforward method of finding the density of ...
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Tags: volume and density application of pressure COSTS associated rectangular gasoline Fluid Statics