Question

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

Answer 1

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