Question

if f= ax + bt^2 + c where f= force x=distance and t = time find the dimensions of a,band c

Answer 1

Given :

• F = ax + bt² + c

To Find :

• Dimensions of a, b and c

Solution :

We're given equation F = ax + bt² + c. Where,

• F is force
• x is displacement
• t is time

And the dimensions formula of following are :

$\: \: \: \: \: \: \: \: \bullet \: \sf{F \: = \: [ML T^{-2}]} \\ \\ \sf{\: \: \: \: \: \: \: \: \bullet \: x \: = \: [M^0 L T^0]} \\ \\ \sf{\: \: \: \: \: \: \: \: \bullet \: t \: = \: [M^0 L^0 T]}$

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$\implies \sf{F \: = \: ax} \\ \\ \implies \sf{[MLT ^{-2}] \: = \: a \: \times \: [M^0 L T^0]} \\ \\ \implies \sf{a \: = \: \dfrac{[MLT^{-2}]}{[M^0 L T^0]}} \\ \\ \implies \sf{a \: = \: [ML^{1 \: - \: 1} T^{-2}]} \\ \\ \implies \sf{a \: = \: [ML ^0 T^{-2}]}$

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$\implies \sf{F \: = \: bt^2} \\ \\ \implies \sf{[MLT^{-2}] \: = \: b \: \times \: [M^0 L^0 T^2]} \\ \\ \implies \sf{b \: = \: \dfrac{[MLT^{-2}]}{[M^0 L^0 T^2]}} \\ \\ \implies \sf{b \: = \: [MLT^{-2 \: - \: 2}]} \\ \\ \implies \sf{b \: = \: [MLT^{-4}]}$

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$\implies \sf{F \: = \: c} \\ \\ \implies \sf{c \: = \: [MLT^{-2}]}$

Answer 2

Explanation:

x+b t^{2}+c x t where F is force, x is distance and t is time. Then what is dimension of a x^{2} c ...