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

The dimensionally consistent equations from the given choices are (b) x=3vt and (h) v=5at, as they both have the same dimensions on each side of the equation.

Step-by-step explanation:

To determine if the given equations are dimensionally consistent, we must ensure that all terms have the same dimensions. We are given the dimensions of s (distance), v (velocity), a (acceleration), and t (time), which are [s] = L, [v] = LT⁻¹, [a] = LT⁻², and [t] = T, respectively. Let’s assess the provided equations:

• (a) a=v / t+xv² / 2 - This is not dimensionally consistent as the dimensions of the two terms v/t and x(v²) are different.
• (b) x=3vt - This is dimensionally consistent as both sides of the equation have the dimension L (since vt has dimension LT⁻¹ × T = L).
• (c) xa²=x²v / t⁴ - This is not dimensionally consistent as the dimensions of xa² and x²v/t⁴ do not match.
• (d) x=vt+vt² / 2 - This is not dimensionally consistent; vt has the dimension L, while vt² does not match this dimension.
• (e) v=x² / at³ - This equation is not dimensionally consistent; the terms x² and at³ on the right do not lead to the dimension LT⁻¹.
• (f) a³=x²v / t⁵ - This is not dimensionally consistent as the dimensions on each side of the equation do not match.
• (g) x=t - This is not dimensionally consistent; they have different dimensions with x as L and t as T.
• (h) v=5at - This equation is dimensionally consistent as both sides result in the dimension LT⁻¹.

Therefore, the only dimensionally consistent equations from the choices given are (b) and (h).

Answer 2

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  =  (x^2 v)/(t^5)`

`B) \   \  x  =  t`

`C \ \ \ v  =  (x^2)/(at^3)`

`D \ \ \ xa^2 = (x^2v)/(t^4)`

`E \ \ \ x  = vt+ (vt^2)/(2)`

`F \ \ \  x = 3vt`

`G \ \ \  v =  5at`

`H \ \ \  a  =  (v)/(t) + (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  =  ((L)/(T^2) )^3 =  L^3\cdot T^(-6)`

and
`(x^2v)/(t^5 ) =  (L^2 L T^(-1))/(T^5)  =  L^3 \cdot T^(-6)`

Hence

`a^3  =  (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

`(x^2 )/(at^3) =  (L^2)/(LT^(-2) T^(3))  =  LT^(-1)`

Hence

`v  =  (x^2)/(at^3)` is dimensionally consistent

Considering D

`xa^2  = L(LT^(-2))^2 =  L^3T^(-4)`

and

`(x^2v)/(t^4)  = (L^2(LT^(-1)))/( T^5) =  L^3 T^(-5)`

Hence

`xa^2 = (x^2v)/(t^4)` is not dimensionally consistent

Considering E

`x =  L`

;

`vt  =  LT^(-1) T =  L`

and

`(vt^2)/(2)  =  LT^(-1)T^(2) =  LT`

Hence

`E \ \ \ x  = vt+ (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)`

,

`(v)/(t)  =  (LT^(-1))/(T)  =  LT^(-2)`

and

`(xv^2)/(2) =  L(LT^(-1))^2 =  L^3T^(-2)`

Hence

`H \ \ \  a  =  (v)/(t) + (xv^2)/(2)` is not dimensionally consistent