A body of mass m = 3.513 ; kg is moving along | vedclass

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(B) The momentum $p$ is given by the product of mass $m$ and velocity $v$: $p = m 	imes v$.Given $m = 3.513 \; kg$ (which has $4$ significant figures

Vedclass · Accepted Answer

$17.6 \; kg \; ms^{-1}$

Vedclass · Answer

(B) The momentum $p$ is given by the product of mass $m$ and velocity $v$: $p = m 	imes v$.Given $m = 3.513 \; kg$ (which has $4$ significant figures) and $v = 5.00 \; ms^{-1}$ (which has $3$ significant figures).Calculating the product: $p = 3.513 	imes 5.00 = 17.565 \; kg \; ms^{-1}$.According to the rules of significant figures,the result of a multiplication should have the same number of significant figures as the measurement with the fewest significant figures.Since $5.00$ has $3$ significant figures,the final result must be rounded to $3$ significant figures.Rounding $17.565$ to $3$ significant figures gives $17.6 \; kg \; ms^{-1}$.

List-$I$ (Number)	List-$II$ (Significant figures)
$A. 1001$	$I. 3$
$B. 010.1$	$II. 4$
$C. 100.100$	$III. 5$
$D. 0.0010010$	$IV. 6$

$A$ body of mass $m = 3.513 \; kg$ is moving along the $x$-axis with a speed of $5.00 \; ms^{-1}$. The magnitude of its momentum is recorded as (Consider significant figures):

Similar Questions

What is the number of significant figures in $11.118 \times 10^{-6} \ V$?

Match List-$I$ with List-$II$.

List-$I$ (Number) List-$II$ (Significant figures)

$A. 1001$ $I. 3$

$B. 010.1$ $II. 4$

$C. 100.100$ $III. 5$

$D. 0.0010010$ $IV. 6$

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