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Maslowich
3 years ago
13

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=v

t+vt2 / 2(e) v=x2 / at3(f) a3=x2v / t5(g) x=t(h) v=5at

Mathematics
1 answer:
Bumek [7]3 years ago
8 0

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

                        A) \   \  a^3  =  \frac{x^2 v}{t^5}

                       

                       B) \   \  x  =  t

 

                       C \ \ \ v  =  \frac{x^2}{at^3}

 

                      D \ \ \ xa^2 = \frac{x^2v}{t^4}

                      E \ \ \ x  = vt+ \frac{vt^2}{2}

                     F \ \ \  x = 3vt

 

                    G \ \ \  v =  5at

 

                    H \ \ \  a  =  \frac{v}{t} + \frac{xv^2}{2}

Generally in dimension

     x - length is represented as  L

     t -  time is represented as T

     m = mass is represented as M

Considering A

           a^3  =  (\frac{L}{T^2} )^3 =  L^3\cdot T^{-6}

and    \frac{x^2v}{t^5 } =  \frac{L^2 L T^{-1}}{T^5}  =  L^3 \cdot T^{-6}

Hence

           a^3  =  \frac{x^2 v}{t^5} is dimensionally consistent

Considering B

            x =  L

and      

            t = T

Hence

      x  =  t  is not dimensionally consistent

Considering C

     v  =  LT^{-1}

and  

    \frac{x^2 }{at^3} =  \frac{L^2}{LT^{-2} T^{3}}  =  LT^{-1}

Hence

   v  =  \frac{x^2}{at^3}  is dimensionally consistent

Considering D

    xa^2  = L(LT^{-2})^2 =  L^3T^{-4}

and

     \frac{x^2v}{t^4}  = \frac{L^2(LT^{-1})}{ T^5} =  L^3 T^{-5}

Hence

    xa^2 = \frac{x^2v}{t^4}  is not dimensionally consistent

Considering E

   x =  L

;

   vt  =  LT^{-1} T =  L

and  

    \frac{vt^2}{2}  =  LT^{-1}T^{2} =  LT

Hence

   E \ \ \ x  = vt+ \frac{vt^2}{2}   is not dimensionally consistent

Considering F

     x =  L

and

    3vt = LT^{-1}T =  L      Note in dimensional analysis numbers are

                                                       not considered

  Hence

       F \ \ \  x = 3vt  is dimensionally consistent

Considering G

    v  =  LT^{-1}

and

    at =  LT^{-2}T =  LT^{-1}

Hence

      G \ \ \  v =  5at   is dimensionally consistent

Considering H

     a =  LT^{-2}

,

       \frac{v}{t}  =  \frac{LT^{-1}}{T}  =  LT^{-2}

and

    \frac{xv^2}{2} =  L(LT^{-1})^2 =  L^3T^{-2}

Hence

    H \ \ \  a  =  \frac{v}{t} + \frac{xv^2}{2}  is not dimensionally consistent

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Note that by letting x+2=\sqrt3\sec t, we are enforcing an invertible substitution which would make it so that t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3} requires 0\le t or \dfrac\pi2. However, \tan t is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which \tan t>0 so that |\tan t|=\tan t. This allows us to write

=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt
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\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C
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\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C

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