It's time for some help for the Scientifically challenged. Today we deal with problems involving dimensional consistency. I will be solving problems from my favorite Physics book, "The Fundamentals of Physics" by Halliday, Resnick and Walker 8th ed.
Chapter 1 problem #12.
To solve this problem we are to considered that all the factors of the numerator are length dimensions so you get a volume dimension as a result of the multiplication of the cube of a length dimension. Dividing the numerator with the denominator, you get a length squared dimension which leads to the final result of a length dimension upon taking the square root.
= (L^2)^1/2
= L
the dimensions are consistent.
Chapter 1 problem #13
First we take note of the dimensions of the quantities called out:
S - length
a - length/length^2, m/s^2
t - time, s
if we were to plug in the values of m = 1 and n =2, the equation would look like this
length = k (length/(time^2))^1 * time^2
with time cancelling out and not considering the dimensionless k, we find
both sides of the equation reflecting length
Dimensional analysis does not give enough info to determine the value of k
Chapter 1 problem #14
Consider the dimensions of the following quantities:
T - time
l - length
g - length/time^2
the quotient inside the radical results in a time^2 quantity. Taking the square root results in both sides of the equation having time units.
Chapter 1 problem #15
Chapter 1 problem #12.
To solve this problem we are to considered that all the factors of the numerator are length dimensions so you get a volume dimension as a result of the multiplication of the cube of a length dimension. Dividing the numerator with the denominator, you get a length squared dimension which leads to the final result of a length dimension upon taking the square root.
r = [ (s-a)(s-b)(s-c)/s]^1/2
= ( L^3/L)^1/2= (L^2)^1/2
= L
the dimensions are consistent.
Chapter 1 problem #13
First we take note of the dimensions of the quantities called out:
S - length
a - length/length^2, m/s^2
t - time, s
if we were to plug in the values of m = 1 and n =2, the equation would look like this
length = k (length/(time^2))^1 * time^2
with time cancelling out and not considering the dimensionless k, we find
both sides of the equation reflecting length
Dimensional analysis does not give enough info to determine the value of k
Chapter 1 problem #14
Consider the dimensions of the following quantities:
T - time
l - length
g - length/time^2
the quotient inside the radical results in a time^2 quantity. Taking the square root results in both sides of the equation having time units.
Chapter 1 problem #15
In the case of equation (a), the left hand side of the equation has length/time dimensions. The right hand side of the equation has length/time dimensions for the first term and length^2/time^2 for the second term. Thus this equation is not dimensionally consistent.
Equation (b) on the other hand has both length dimensions on either side of the equation making it dimensionally consistent.
Chapter 1 problem #16
Force has the units of kg•m/s^2. Multiplying this with r^2, and dividing this with kg^2, you get m^3/kg•s^2 for the units of G.
Chapter 1 problem #17
We first define the first term coefficient A = 1.50 Mft^3/mos. Converting to ft^3/s we go through the following:
A = 1.5x10^6ft^3/mos x 1mo/30d x 1d/24h x 1h/60m x 1min/60s = 0.579 ft^3/s
in a similar fashion
B = 0.008x10^6 ft^3/mos^2 x (1mon/30d)^2 x (1d/24h)^2 x (1h/60min)^2 x (1min/60s )
= 0.19 x 10^-9 ft^3/s^2
V = 0.58 ft^3/s t + 1.19 x 10^-9 ft^3/s^2 t^2
