Question

Answer it fraands

Equal forces f acts on isolated bodies A and B.The mass of B is three times that of A .The magnitude of the acceleration of A is?

Answer 1

Step-by-step explanation:

this will be your right answer

Answer 2

Answer:

The answer to the given question is the magnitude of acceleration of A is 3 times that of B

Step-by-step explanation:

Given:

Two bodies A and B where equal forces ( f ) acts on them.

The mass of B of three times of A.

To find :

The magnitude of the acceleration of A is

Solution :

let us assume the force is 15x N

And the mass of A is 5y kg.

The mass of B is 3 times of A which is 3(5y)=15y kg.

The formula of the force is

F= ma

a= F/m.

The acceleration of B is

$= \frac{15x}{15y} = \frac{x}{y} \frac{m}{ {s}^{2} }$

The acceleration of A is

$\frac{15x}{5y} = \frac{3x}{y} \frac{m}{ {s}^{2} }$

The division of acceleration of A by the acceleration of B is

$\frac{ \frac{3x}{y} }{ \frac{x}{y} }$

on cancelling the similar terms we get the remaining value as 3

Therefore, the final value obtained is 3.

$\frac{acceleration \: of \: a}{acceleration \: of \: b} = 3 \\ acceleration \: of \: a= 3 \times acceleration \: of \: b$

Therefore, the acceleration of A is equal to the 3 times the acceleration of B.

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