1. If the change in position delta(x) is related to velocity v (w/ units of m/s) in the equation delta(x)=Av, the constant A has which dimension?
a. m/s^2
b. m
c. s
d. m^2
2. If a is acceleration (m/s^2), delta(v) is change in velocity (m/s), delta(x) is change in position (m), and delta(t) is the time interval (s), which equation is NOT dimensionally correct?
a. delta(t) = delta(x)/v
b. a = v^2/delta(x)
c. delta(v)=a/delta(t)
d. delta(t)^2=2delta(x)/a
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With number 1, I'm just not really sure. I think the unit for delta(x) is throwing me off a little.
For number 2, I know what it's asking, but I'm not too good at finding dimensionally correct equations. I'm leaning toward d, however. I don't see time getting squared. A little help, please?
Thanks!
Let's break down both questions step by step:
1. If the change in position delta(x) is related to velocity v in the equation delta(x) = Av, the constant A has which dimension?
To determine the dimensions of A, we need to analyze the equation. The change in position, delta(x), is measured in meters (m), and the velocity, v, is measured in meters per second (m/s). To get the dimensions of A, we divide the dimensions of delta(x) by the dimensions of v.
Dimensions of delta(x): [L]
Dimensions of v: [L]/[T]
Therefore, the dimensions of A will be [L]/([L]/[T]) = [L] * [T]/[L] = [T].
So, the constant A has the dimension of time.
Answer: c. s
2. If a is acceleration (m/s^2), delta(v) is change in velocity (m/s), delta(x) is change in position (m), and delta(t) is the time interval (s), which equation is NOT dimensionally correct?
To determine which equation is NOT dimensionally correct, we can compare the dimensions on both sides of each equation.
a. delta(t) = delta(x)/v
Dimensions of delta(t): [T]
Dimensions of delta(x): [L]
Dimensions of v: [L]/[T]
Dimensions on the left side: [T]
Dimensions on the right side: [L] / [L]/[T] = [T]
b. a = v^2 / delta(x)
Dimensions of a: [L] / [T]^2
Dimensions of v^2 : ([L]/[T])^2 = [L]^2 / [T]^2
Dimensions of delta(x): [L]
Dimensions on the left side: [L] / [T]^2
Dimensions on the right side: [L]^2 / [T]^2 / [L] = [L] / [T]^2
c. delta(v) = a / delta(t)
Dimensions of delta(v): [L]/[T]
Dimensions of a: [L] / [T]^2
Dimensions of delta(t): [T]
Dimensions on the left side: [L]/[T]
Dimensions on the right side: [L] / [T]^2 / [T] = [L] / [T]^2
d. delta(t)^2 = 2delta(x) / a
Dimensions of delta(t)^2: [T]^2
Dimensions of 2delta(x): 2[L]
Dimensions of a: [L] / [T]^2
Dimensions on the left side: [T]^2
Dimensions on the right side: [L] / [T]^2 / [L] = [L] / [T]^2
Comparing the dimensions on both sides of each equation, we find that equation d does not have consistent dimensions.
Answer: d. delta(t)^2 = 2delta(x) / a is NOT dimensionally correct.
For question 1, we are given the equation delta(x) = Av, where delta(x) represents the change in position and v represents velocity. We need to determine the dimension of the constant A.
To find the dimension of A, we need to analyze the units on both sides of the equation. The unit of delta(x) is meters (m) and the unit of v is meters per second (m/s). Therefore, the dimension of A must cancel out the seconds in the denominator of v, leaving us with just meters for delta(x).
From the given options, we find that only option b, m, matches this requirement and cancels out the seconds in the denominator. Therefore, the correct answer for question 1 is b. m.
For question 2, we need to determine which equation is NOT dimensionally correct.
a. delta(t) = delta(x)/v:
On the left-hand side, delta(t) has the unit of seconds (s). On the right-hand side, delta(x) has the unit of meters (m), and v has the unit of meters per second (m/s). Dividing meters by meters per second cancels out the meters in the numerator and denominator, leaving just seconds. Therefore, this equation is dimensionally correct.
b. a = v^2/delta(x):
On the left-hand side, a represents acceleration and has the unit of meters per second squared (m/s^2). On the right-hand side, v squared has the unit of (m/s)^2, and delta(x) has the unit of meters (m). Dividing (m/s)^2 by meters cancels out the meters in the numerator and denominator, leaving just (m/s)^2. Therefore, this equation is dimensionally correct.
c. delta(v) = a/delta(t):
On the left-hand side, delta(v) represents change in velocity and has the unit of meters per second (m/s). On the right-hand side, a represents acceleration and has the unit of meters per second squared (m/s^2), and delta(t) has the unit of seconds (s). Dividing meters per second squared by seconds cancels out the seconds in the denominator, leaving just meters per second. Therefore, this equation is dimensionally correct.
d. delta(t)^2 = 2delta(x)/a:
On the left-hand side, delta(t) squared has the unit of seconds squared (s^2). On the right-hand side, 2delta(x) has the unit of meters (m), and a represents acceleration and has the unit of meters per second squared (m/s^2). Dividing meters by meters per second squared cancels out the meters in the numerator and denominator, leaving just seconds squared. Therefore, this equation is dimensionally correct.
None of the equations is NOT dimensionally correct. Therefore, the correct answer for question 2 is none of the options.
on the first deltax has units of meters.
on the second. Put the dang units for each in, and just reduce to see if it matches.
For instance:
a. s=m/(m/s)= s
duh, those match.
now do the rest