Question

1. complete the following table in the form xm^n l^p t^q

(table: quantity, common units, m, l, t columns:
mass: kg
length: m
time: s
area: m²
volume: l
gravity: m/s²
velocity: mi/h, m/s
acceleration: m/s²
volumetric flow rate: m³/s
height: m
power: kg·m²/s³
energy: kg·m²/s²
force: kg·m/s²
density: kg/m³

1. identify the following quantities through the use of fundamental dimensions. choose from the following:

acceleration, energy, power, velocity, force
example: mass · gravity = kg · m/s² = kg·m/s² (force)

a) power / (mass · gravity)
b) density · volumetric flow rate · gravity · height
c) mass · gravity · height
d) force / mass

Answer:

Part 3: Completing the Table (Dimension Analysis)

To complete the table, we use the fundamental dimensions: \( M \) (mass), \( L \) (length), \( T \) (time). We analyze each quantity's units to express them in terms of \( M^a L^b T^c \).

Step 1: Mass

• Unit: \( \text{kg} \)
• Dimensions: \( M^1 L^0 T^0 \) (mass is a fundamental dimension, so only \( M \) with exponent 1, others 0)

Step 2: Length

• Unit: \( \text{m} \)
• Dimensions: \( M^0 L^1 T^0 \) (length is fundamental, so \( L \) with exponent 1)

Step 3: Time

• Unit: \( \text{s} \)
• Dimensions: \( M^0 L^0 T^1 \) (time is fundamental, so \( T \) with exponent 1)

Step 4: Area

• Unit: \( \text{m}^2 \) (length squared)
• Dimensions: \( M^0 L^2 T^0 \) (since area = length × length, \( L^2 \))

Step 5: Volume

• Unit: \( \text{L} \) (or \( \text{m}^3 \), since 1 L = \( 10^{-3} \text{m}^3 \))
• Dimensions: \( M^0 L^3 T^0 \) (volume = length × length × length, \( L^3 \))

Step 6: Gravity (Acceleration)

• Unit: \( \text{m/s}^2 \) (length per time squared)
• Dimensions: \( M^0 L^1 T^{-2} \) (acceleration = length / time², so \( L T^{-2} \))

Step 7: Velocity

• Unit: \( \text{m/s} \) (or \( \text{mi/h} \), but fundamental dimensions use \( \text{m/s} \))
• Dimensions: \( M^0 L^1 T^{-1} \) (velocity = length / time, so \( L T^{-1} \))

Step 8: Acceleration

• Unit: \( \text{m/s}^2 \) (same as gravity, since acceleration is length per time squared)
• Dimensions: \( M^0 L^1 T^{-2} \)

Step 9: Volumetric Flow Rate

• Unit: \( \text{m}^3/\text{s} \) (volume per time)
• Dimensions: \( M^0 L^3 T^{-1} \) (volume is \( L^3 \), divided by time \( T \), so \( L^3 T^{-1} \))

Step 10: Height

• Unit: \( \text{m} \) (same as length)
• Dimensions: \( M^0 L^1 T^0 \)

Step 11: Power

• Unit: \( \text{kg·m}^2/\text{s}^3 \) (mass × length² / time³)
• Dimensions: \( M^1 L^2 T^{-3} \) (from \( \text{kg} = M \), \( \text{m}^2 = L^2 \), \( \text{s}^3 = T^3 \), so \( M L^2 T^{-3} \))

Step 12: Energy

• Unit: \( \text{kg·m}^2/\text{s}^2 \) (mass × length² / time²)
• Dimensions: \( M^1 L^2 T^{-2} \) (from \( \text{kg} = M \), \( \text{m}^2 = L^2 \), \( \text{s}^2 = T^2 \), so \( M L^2 T^{-2} \))

Step 13: Force

• Unit: \( \text{kg·m}/\text{s}^2 \) (mass × length / time²)
• Dimensions: \( M^1 L^1 T^{-2} \) (from \( \text{kg} = M \), \( \text{m} = L \), \( \text{s}^2 = T^2 \), so \( M L T^{-2} \))

Step 14: Density

• Unit: \( \text{kg}/\text{m}^3 \) (mass / volume)
• Dimensions: \( M^1 L^{-3} T^0 \) (mass \( M \) divided by volume \( L^3 \), so \( M L^{-3} \))

Part 4: Identifying Quantities via Fundamental Dimensions

We analyze each option by substituting the dimensions of each quantity and simplifying.

Option A: \( \frac{\text{power}}{\text{mass} \cdot \text{gravity}} \)

• Power dimensions: \( M L^2 T^{-3} \)
• Mass dimensions: \( M \)
• Gravity dimensions: \( L T^{-2} \)
• Substitute: \( \frac{M L^2 T^{-3}}{M \cdot L T^{-2}} = \frac{L^2 T^{-3}}{L T^{-2}} = L T^{-1} \) (which is velocity, since velocity has dimensions \( L T^{-1} \))

Option B: \( \text{density} \cdot \text{volumetric flow rate} \cdot \text{gravity} \cdot \text{height} \)

• Density: \( M L^{-3} \)
• Volumetric flow rate: \( L^3 T^{-1} \)
• Gravity: \( L T^{-2} \)
• Height: \( L \)
• Multiply: \( (M L^{-3}) \cdot (L^3 T^{-1}) \cdot (L T^{-2}) \cdot (L) = M \cdot (L^{-3 + 3 + 1 + 1}) \cdot (T^{-1 - 2}) = M L^2 T^{-3} \) (which is power, since power has dimensions \( M L^2 T^{-3} \))

Option C: \( \text{mass} \cdot \text{gravity} \cdot \text{height} \)

• Mass: \( M \)
• Gravity: \( L T^{-2} \)
• Height: \( L \)
• Multiply: \( M \cdot L T^{-2} \cdot L = M L^2 T^{-2} \) (which is energy, since energy has dimensions \( M L^2 T^{-2} \))

Option D: \( \frac{\text{force}}{\text{mass}} \)

• Force: \( M L T^{-2} \)
• Mass: \( M \)
• Divide: \( \frac{M L T^{-2}}{M} = L T^{-2} \) (which is acceleration, since acceleration has dimensions \( L T^{-2} \))

Final Table (Part 3)

Quantity	Common units	\( M \)	\( L \)	\( T \)
length	m	0	1	0
time	s	0	0	1
area	\( \text{m}^2 \)	0	2	0
volume	L (or \( \text{m}^3 \))	0	3	0
gravity	\( \text{m/s}^2 \)	0	1	-2
velocity	\( \text{m/s} \)	0	1	-1
acceleration	\( \text{m/s}^2 \)	0	1	-2
volumetric flow rate	\( \text{m}^3/\text{s} \)	0	3	-1
height	m	0	1	0
power	\( \text{kg·m}^2/\text{s}^3 \)	1	2	-3
energy	\( \text{kg·m}^2/\text{s}^2 \)	1	2	-2
force	\( \text{kg·m}/\text{s}^2 \)	1	1	-2
density	\( \text{kg}/\text{m}^3 \)	1	-3	0

Part 4 Answers

• A) \( \frac{\text{power}}{\text{mass} \cdot \text{gravity}} \): Velocity
• B) \( \text{density} \cdot \text{volumetric flow rate} \cdot \text{gravity} \cdot \text{height} \): Power
• C) \( \text{mass} \cdot \text{gravity} \cdot \text{height} \): Energy
• D) \( \frac{\text{force}}{\text{mass}} \): Acceleration

Final Answers (Part 4)

A) Velocity
B) Power
C) Energy
D) Acceleration