Which of the following are dimensionally consistent? (Choose all that apply.)(a) a=v / t+xv2 / 2(b) x=3vt(c) xa2=x2v / t4(d) x=vt+vt2 / 2(e) v=x2 / at3(f) a3=x2v / t5(g) x=t(h) v=5at
Answers
Complete Question
The complete question is shown on the first uploaded image
Answer:
A
is dimensionally consistent
B
is not dimensionally consistent
C
is dimensionally consistent
D
is not dimensionally consistent
E
is not dimensionally consistent
F
is dimensionally consistent
G
is dimensionally consistent
H
is not dimensionally consistent
Step-by-step explanation:
From the question we are told that
The equation are
[tex]A) \ \ a^3 = \frac{x^2 v}{t^5}[/tex]
[tex]B) \ \ x = t [/tex]
[tex]C \ \ \ v = \frac{x^2}{at^3}[/tex]
[tex]D \ \ \ xa^2 = \frac{x^2v}{t^4}[/tex]
[tex]E \ \ \ x = vt+ \frac{vt^2}{2}[/tex]
[tex]F \ \ \ x = 3vt[/tex]
[tex]G \ \ \ v = 5at[/tex]
[tex]H \ \ \ a = \frac{v}{t} + \frac{xv^2}{2}[/tex]
Generally in dimension
x - length is represented as L
t - time is represented as T
m = mass is represented as M
Considering A
[tex]a^3 = (\frac{L}{T^2} )^3 = L^3\cdot T^{-6}[/tex]
and [tex]\frac{x^2v}{t^5 } = \frac{L^2 L T^{-1}}{T^5} = L^3 \cdot T^{-6}[/tex]
Hence
[tex]a^3 = \frac{x^2 v}{t^5}[/tex] is dimensionally consistent
Considering B
[tex]x = L[/tex]
and
[tex]t = T[/tex]
Hence
[tex]x = t[/tex] is not dimensionally consistent
Considering C
[tex]v = LT^{-1}[/tex]
and
[tex]\frac{x^2 }{at^3} = \frac{L^2}{LT^{-2} T^{3}} = LT^{-1}[/tex]
Hence
[tex]v = \frac{x^2}{at^3}[/tex] is dimensionally consistent
Considering D
[tex]xa^2 = L(LT^{-2})^2 = L^3T^{-4}[/tex]
and
[tex]\frac{x^2v}{t^4} = \frac{L^2(LT^{-1})}{ T^5} = L^3 T^{-5}[/tex]
Hence
[tex] xa^2 = \frac{x^2v}{t^4}[/tex] is not dimensionally consistent
Considering E
[tex]x = L[/tex]
;
[tex]vt = LT^{-1} T = L[/tex]
and
[tex]\frac{vt^2}{2} = LT^{-1}T^{2} = LT[/tex]
Hence
[tex]E \ \ \ x = vt+ \frac{vt^2}{2}[/tex] is not dimensionally consistent
Considering F
[tex]x = L[/tex]
and
[tex]3vt = LT^{-1}T = L[/tex] Note in dimensional analysis numbers are
not considered
Hence
[tex]F \ \ \ x = 3vt[/tex] is dimensionally consistent
Considering G
[tex]v = LT^{-1}[/tex]
and
[tex]at = LT^{-2}T = LT^{-1}[/tex]
Hence
[tex]G \ \ \ v = 5at[/tex] is dimensionally consistent
Considering H
[tex]a = LT^{-2}[/tex]
,
[tex]\frac{v}{t} = \frac{LT^{-1}}{T} = LT^{-2}[/tex]
and
[tex]\frac{xv^2}{2} = L(LT^{-1})^2 = L^3T^{-2}[/tex]
Hence
[tex]H \ \ \ a = \frac{v}{t} + \frac{xv^2}{2}[/tex] is not dimensionally consistent
We want to see which ones of the given expressions are dimensionally consistent. We will see that the correct options are:
a) x = 3*v*th) v = 5*a*tWhat means to be dimensionally consistent?
This means that we have the same units in the left and in the right side of the equation.
The units are:
a = [m/s^2]x = [m]v = [m/s]t = [s]Now we can analyze the expressions to see the units in each one, I will show you how to do it:
a) a = v/t + x*v^2
Replacing the units we have:
[m/s^2] = [m/s]/[s] + [m]*[m^2/s^2]
[m/s^2] = [m/s^2] + [m^3/s^2]
You can see that we have an m^3 in the right side, so these are not equivalent.
b) x = 3*v*t
Replacing the units we have:
[m] = 3*[m/s]*[s] = 3*[m]
So yes, the units are the same in both sides, so this is dimensionally consistent.
With the same procedure we can see that:
c) [m^3/s^2] = [m^3/s] not consistentd) [m] = [m] + [m*s] not consistente) [m/s] = [m^2] not consistentf) [m^3/s^6] = [m^3/s] not consistentg) [m] = [s] not consistenth) [m/s] = 5*[m/s] consistentSo the correct options are b and h.
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Step-by-step explanation:
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Answers
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Answers
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Step-by-step explanation:
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Answer:
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Answer:
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you just multiply.