The dimensional consistency of the terms can be used to detect algebraic errors. Consider the following equation:FA=gA+ρv2+maL2where v is a velocity's magnitude, a is an acceleration's magnitude, A is an area, m is a mass, L is a length (or distance), g is another acceleration's magnitude, ρ is a density (mass per unit volume), and F
is a force's magnitude. Which terms are dimensionally homogeneous?
Check all that apply. I know the answers are1 2 and 4 but don't know why. explain so I can understand.
1) ρv^2
2) ma/L^2
3) g/A
4) F/A
/ A + ρ v^2 +m a / L^2
F/A = Newtons/ m^2 = kg m/s^2 / m^2 = kg / m s^2
m a/L^2 = kg m/s^2 / m^2 = kg / m s^2 same
g / A = m/s^2 / m^2 = 1 / m s^2 NO KILOGRAMS !
ρ v^2 = kg/m^3 * m^2/s^2 = kg / m s^2 same
equation: /A+ ρv^2 + ma/L^2
Ah, the wonderful world of dimensional consistency! Let's break it down to understand which terms are dimensionally homogeneous.
1) ρv^2: The first term, ρv^2, consists of density (ρ) multiplied by velocity squared (v^2). The units of density are mass per unit volume (mass/volume), and the units of velocity squared are (distance/time)^2. When we multiply these units together, we get (mass/volume) * (distance/time)^2. Simplifying, we have mass * distance^2 / (volume * time^2). Thus, this term is not dimensionally homogeneous. Oops!
2) ma/L^2: The second term, ma/L^2, consists of mass (m) multiplied by acceleration (a), and divided by length squared (L^2). The units of mass are simply mass, the units of acceleration are (distance/time^2), and length squared has the units (distance)^2. Multiplying these units together, we have mass * distance / (time^2 * distance^2), which simplifies to mass / (time^2 * distance). Therefore, this term is dimensionally homogeneous. Hooray!
3) g/A: The third term, g/A, involves acceleration (g) divided by area (A). The units of acceleration are (distance/time^2), and the units of area are (distance)^2. Dividing these units, we get (distance/time^2) / (distance^2), which simplifies to 1 / (time^2 * distance). This term is also dimensionally homogeneous. Woohoo!
4) F/A: Finally, the fourth term F/A represents force (F) divided by area (A). The units of force are mass * acceleration, and the units of area are (distance)^2. Dividing these units, we have (mass * acceleration) / (distance^2), which is perfectly dimensionally homogeneous. Yippee!
To summarize, the dimensionally homogeneous terms in the given equation are:
1) ma/L^2
2) g/A
3) F/A
I hope this explanation helped clear things up and added a touch of humor to your learning experience!
To determine which terms are dimensionally homogeneous, we need to verify if all the terms have the same dimensions.
1) ρv^2:
The dimension for density (ρ) is mass/volume, and the dimension for velocity squared (v^2) is (length/time)^2. Multiplying these dimensions together, we get (mass/volume) * (length^2/time^2). This is the dimension for force, which means ρv^2 is dimensionally homogeneous.
2) ma/L^2:
The dimension for mass (m) is mass, the dimension for acceleration (a) is length/time^2, and the dimension for length squared (L^2) is length^2. Dividing the dimension of mass by the dimension of length squared, we get mass / length^2. This is the dimension for pressure or stress, not force. Therefore, ma/L^2 is not dimensionally homogeneous.
3) g/A:
The dimension for acceleration (g) is length/time^2, and the dimension for area (A) is length^2. Dividing the dimension of length/time^2 by the dimension of length^2, we get 1/time^2. This is the dimension for frequency, not force. Therefore, g/A is not dimensionally homogeneous.
4) F/A:
The dimension for force (F) is mass * length / time^2, and the dimension for area (A) is length^2. Dividing the dimension of mass * length / time^2 by the dimension of length^2, we get mass / (length * time^2). This is the dimension for pressure or stress, not force. Therefore, F/A is not dimensionally homogeneous.
Therefore, the terms that are dimensionally homogeneous are:
1) ρv^2
2) ma/L^2
I hope this explanation helps you understand why these terms are dimensionally homogeneous!
To determine which terms are dimensionally homogeneous in the given equation, we need to analyze the dimensions of each term and check if they are consistent.
Dimensional homogeneity is a property that indicates all terms in an equation have the same physical dimensions. In other words, the units on both sides of the equation must be the same.
Let's break down each term and analyze their dimensions:
1) ρv^2:
- ρ represents density, which has the dimensions of mass per unit volume (M/L^3).
- v^2 represents velocity squared, which has the dimensions of L^2/T^2 (length squared divided by time squared).
Since mass per unit volume times length squared divided by time squared gives dimensions of mass, the first term is dimensionally homogeneous.
2) ma/L^2:
- m represents mass, which has the dimensions of mass (M).
- a represents acceleration's magnitude, which has the dimensions of L/T^2 (length divided by time squared).
- L^2 represents length squared, which has the dimensions of L^2.
Since mass times length divided by time squared divided by length squared gives dimensions of mass per unit length, the second term is dimensionally homogeneous.
3) g/A:
- g represents acceleration's magnitude, which has the dimensions of L/T^2 (length divided by time squared).
- A represents area, which has the dimensions of L^2.
Since length divided by time squared divided by length squared gives dimensions of 1/length, the third term is not dimensionally homogeneous.
4) F/A:
- F represents force's magnitude, which has the dimensions of M·L/T^2 (mass times length divided by time squared).
- A represents area, which has the dimensions of L^2.
Since force's magnitude divided by area gives dimensions of pressure (M/L·T^2), the fourth term is dimensionally homogeneous.
Therefore, the dimensionally homogeneous terms in the given equation are 1) ρv^2, 2) ma/L^2, and 4) F/A. These terms have consistent physical dimensions and can be used to check for algebraic errors when inspecting the equation.