3.The volume of a sphere is given by v=4/3pir^3 where r is the radius of the sph

Question

3.The volume of a sphere is given by v=4/3pir^3 where r is the radius of the sphere. what is the density in g/mL of a sphere which weighs 2.00lbs and has a radius of 3.50 inches?

7. Consider a stone column 3.00 ft in diameter and 10.00 ft high. What is its mass in pounds if its density is 3.50 g/mL? Show set-up and the results of the calculations.

9. Suppose that a student performed the experiment and calculations perfectly as directed except that, unknown to the student, the balance was not zeroed but weighed 0.100 too high throughout the experiment. Would the calculated value for the density of the unknown liquid be correct? That is, would it be equal to, higher, or lower than the true value. Why?

10. If the same balance that was used in question 9 was used to determine the mass of the metal cylinder, would the calculated value be equal, higher, or lower than the true value? Why?

Explanation / Answer

3.The volume of a sphere is given by v=4/3pir^3 where r is the radius of the sphere. what is the density in g/mL of a sphere which weighs 2.00lbs and has a radius of 3.50 inches?

Solution :-

The formula to calculate the volume of the sphere is given that is v=4/3pir^3

Lets calculate the volume of the sphere using the given radius

Need to convert the radius from inches to centimeter

3.50 inch * 2.54 cm/ 1 inch = 8.89 cm

v=4/3pir^3

v= 4/3 * 3.14 * (8.89 cm)^3

v= 2942 cm3

1 cm3 = 1 ml

Therefore volume of sphere = 2942 ml

now lets convert the mass of sphere from lb to gram

2.00 lb * 453.592 g / 1 lb = 907.184 g

Now   lets calculate the density

Density = mass / volume

              = 907.184 g / 2942 ml

              = 0.308 g /ml

Therefore density of sphere = 0.308 g/ml

7. Consider a stone column 3.00 ft in diameter and 10.00 ft high. What is its mass in pounds if its density is 3.50 g/mL? Show set-up and the results of the calculations.

Solution :-

Lets first calculate the volume of the column

Radius = diameter / 2

             = 3.00 /2

            = 1.50 ft

Formula , v= h*pi*r^2

                   = 10.0 ft*3.14*(1.50ft)^2

                    = 70.65 ft3

Now lets convert ft3 to cm3

1 ft3 = 28316.85 cm3

70.65 ft3 * 28316.85 cm3 / 1 ft3 = 2000585 cm3

Now lets calculate the mass using the volume and density

Mass = volume * density

         = 2000585 ml*3.50 g per ml

         = 7002047 g

Now lets convert gram to pounds

7002047 g * 1 lb / 453.592 g = 15437 lb

9. Suppose that a student performed the experiment and calculations perfectly as directed except that, unknown to the student, the balance was not zeroed but weighed 0.100 too high throughout the experiment. Would the calculated value for the density of the unknown liquid be correct? That is, would it be equal to, higher, or lower than the true value. Why?

Solution :- Since the balance was not set to zero and it weighs 0.100 too high throughout therefore this will add up in the mass of the liquid and since the density is nothing but the mass to volume ratio therefore the calculated density would be higher.

10. If the same balance that was used in question 9 was used to determine the mass of the metal cylinder, would the calculated value be equal, higher, or lower than the true value? Why?

Solution :- If the same balance is used to determine the mass of the metal cylinder then the mass of the metal cylinder would be higher than the true value by 0.100 amount.because balance is not set to zero and it already weighs 0.100.

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