PHY 2048 Homework for Module 1 (weekly assignment, Optional) Dimensions of Physical Quantities Flashcards


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1

In classical mechanics there are three base dimensions. Length is one of them. What are the other two?

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mass and time

2

Find the dimensions [V] of volume.Express your answer as powers of length (l), mass (m), and time (t).

[V] =l3

3

Find the dimensions [v] of speed.Express your answer as powers of length (l), mass (m), and time (t).

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[V] = l/t

4

Find the dimensions [a] of acceleration.Express your answer as powers of length (l), mass (m), and time (t).

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[a] = l/t2

5

How many nanoseconds does it take light to travel 4.00 ft in vacuum? Express your answer in nanoseconds.

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t= 4.08 ns

6

The density of uranium is 18.8 g/cm3. What is this value in kilograms per cubic meter? Express your answer in kilograms per cubic meter.

ρ =1.88×104kg/m3

18800 kg/m^3 This is a simple matter of multiplying by the appropriate conversion factors. So let's see what they are: We have grams and want kilograms. That will be a simple matter of dividing by 1000. We have cm^2 and want m^3. This is slightly more difficult. There's 100 cm per meter. And since we need to cube it, that will be 100^3 = 1000000. So the result will be 18.8 g/cm^3 * 1000000 cm^3/m^3 / 1000 g/kg = 18800 kg/m^3

7

The volume of a solid cylinder is given by V=πr2h, where r is the radius and ℎ is the height. You measure the radius and height of a thin cylindrical wire and obtain the results r = 0.036 cmcm and ℎ = 12.1 cm. What do your measurements give for the volume of the wire in mm3mm3? Use the correct number of significant figures in your answer.Express your answer in cubic millimeters.

V =49 mm3

Given data , Radius of cylinder (r)=0.036 cmHeight of the cylinder ( h)=12.1 cm

Unit of the Volume is cubic meters or cubic centimeters and cubic millimeters.

The volume of a solid cylinder is given by V=πr^2h where r is the radius and h is the height. Substituting the given values in the above formula and we get. V=π×(0.036cm)^2×(12.1cm) V=0.049cm^3=0.049×1000 mm^3 V=49mm^3.

We know that, 1cm^3 =1×1000mm^3.

Correct answer is 49 mm^3.

8

In the fall of 2002, a group of scientists at Los Alamos National Laboratory determined that the critical mass of neptunium-237 is about 60 kg. The critical mass of a fissionable material is the minimum amount that must be brought together to start a chain reaction. This element has a density of 19.5 g/cm3. What would be the radius of a sphere of this material that has a critical mass? Express your answer in centimeters.

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r = 9.01 cm