Question

A particle is moving on a straight line with a speed of about 0.5 m/s. To measure its speed, points are marked on the line of motion, which are exactly separated by 5 mm. To have an accuracy of 99% in the determination of speed, the minimum least count of the measuring clock is nx 10-4 s. Find value of n.​

Answer 1

Answer:

15m/s

hope you under stand

Answer 2

Given:

v=0.5 m/s

Find value of n=?

$\mathrm{t}=\frac{\mathrm{d}}{\mathrm{V}}$

$=\frac{5 \times 10^{-3}}{0.5}$

$=10 \times 10^{-3}$

$=10^{-4}$

$\frac{\Delta t}{t}=99 \%$

$\frac{\Delta t}{t}=\frac{99}{100}$

$\Delta t=\frac{99}{100} \times 10^{-2}$

$=99\times10^{-4} s e c$

so here n = 99

Hence the value of n is 99 or  $99\times10^{-4} s e c$