Question

please solve this fast...​

Answer 1

Answer:

• The dimensions of a is L T⁻²
• The dimensions of b is L
• The dimensions of c is T

Given:

1. $\sf v=at+\dfrac{b}{t+c}$

Explanation:

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From principle of homogeneity we know that same quantities can only be added,therefore the dimensional formula of v should be equal to at and $\textsf{\textbf{ \dfrac{\text b}{\text{t + c}} }}$ individually.

Now, lets find the dimension of "a"

$\displaystyle\longrightarrow\sf \Big[v\Big]=\Big[a\Big]\;\Big[t\Big]\\\\\\\longrightarrow\sf \bigg[L\;T^{-1}\bigg]=\Big[a\Big]\;\Big[T\Big]\\\\\\\longrightarrow\sf \Big[a\Big]=\dfrac{\bigg[L\;T^{-1}\bigg]}{\Big[T\Big]}\\\\\\\longrightarrow\sf \Big[a\Big] = \bigg[L\;T^{-1}\bigg]\; \Big[T^{-1}\Big]\\\\\\\longrightarrow\sf \Big[a\Big]=\bigg[L\;T^{-2}\bigg]\\\\\\\longrightarrow \large{\underline{\boxed{\textsf{\textbf{\Big[a\Big]=\bigg[L\;T ^{\text{ -2}} \bigg]}}}}}$

∴ The dimensions of a is L T⁻².

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Now lets find dimensions of "c"

According to principle,

$\displaystyle\longrightarrow\sf \Big[c\Big]=\Big[t\Big]\\\\\\\longrightarrow\sf \Big[c\Big]=\Big[T\Big]\\\\\\\longrightarrow \large{\underline{\boxed{\textsf{\textbf{\Big[c\Big]=\Big[T\Big]}}}}}$

∴ The dimensions of c is T.

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Now lets find dimensions of "b",

$\displaystyle\longrightarrow\sf \Big[v\Big]=\dfrac{\Big[b\Big]}{\Big[t\Big]}\\\\\\\longrightarrow\sf \Big[LT^{-1}\Big]=\dfrac{\Big[b\Big]}{\Big[T\Big]}\\\\\\\longrightarrow\sf \Big[b\Big]=\Big[LT^{-1}\Big]\;\Big[T\Big]\\\\\\\longrightarrow\sf \Big[b\Big]=\Big[L\Big]\\\\\\\longrightarrow \large{\underline{\boxed{\textsf{\textbf{\Big[b\Big]=\Big[L\Big]}}}}}$

∴ The dimensions of b is L.

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Answer 2

$Questions:-$

• $If \: \: velocity \: \: v \: \: = \: \: {at} \: + \: \frac{b}{t \: + \: c} \: \: where \: \: \: t ⟹ \: \: time \: \: and \: \: a, \: \: b, \: \: c \: \: ⟹ constants \: \: then \: \: find \: \: the \: \: dimensions \: \: of \: \: a, \: \: b, \: \: and \: \: c.$

$Given:-$

• $velocity \: \: v \: \: = \: \: {at} \: + \: \frac{b}{t \: + \: c}$

• $a, \: \: b, \: \: c \: \: = constants$

$To \: \: Find:-$

• $find \: \: dimensions \: \: of \: \: a, \: \: b, \: \: and \: \: c.$

$Solutions:-$

$\: \: \: \: \: {[v]} \: \: = \: \: {[at]} \: \: + \: \: {[\frac{b}{t \: + \: c}]}$

$\: \: \: \: \: {[v]} \: \: = \: \: {[at]} \: \: = \: \: {[\frac{b}{t \: + \: c}]}$

$\: \: \: \: \: {[a]} \: \: = \: \: \frac{[v]}{[t]} \: \: = \: \: \frac{[{L}^{1} \: \: {T}^{-1}]}{[{T}^{1}]} \: \: = \: \: {[{L}^{1} \: \: {T}^{-2}]}$

$The \: \: dimensions \: \: a \: \: is \: \: {[{L} \: \: {T}^{-2}]}$

$\: \: \: \: \: {[t \: + \: c]} \: \: = \: \: {[T]} \: \: = \: \: {[c]}$

$\: \: \: \: \: {[c]} \: \: = \: \: {[T]}$

$The \: \: dimensions \: \: c \: \: is \: \: {[T]}$

$\: \: \: \: \: {[v]} \: \: = \: \: {[\frac{b}{t \: + \: c}]}$

$\: \: \: \: \: {[b]} \: \: = \: \: {[v]} \: × \: {[{T}^{1}]}$

$\: \: \: \: \: {[b]} \: \: = \: \: {[{L}^{1} \: \: {T}^{-1}]} \: × \: {[{T}^{1}]}$

$\: \: \: \: \: {[b]} \: \: = \: \: {[{L}^{1}]}$

$The \: \: dimensions \: \: b \: \: is \: \: {[{L}^{1}]}$

$\: \: \: \: \: ⟹ a \: \: = \: \: {[{L} \: \: {T}^{-2}]}$

$\: \: \: \: \: ⟹ b \: \: = \: \: {[L]}$

$\: \: \: \: \: ⟹ c \: \: = \: \: {[T]}$

$So,$

$\: \: \: \: \: the \: \: dimensions \: \: of \: \: a, \: \: b, \: \: and \: \: c \: \: is \: \: {[{L} \: \: {T}^{-2}]}, \: \: {[L]}, \: \: and \: \: {[T]}$

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