Physics, asked by bpsrivastava1973, 9 months ago

a farmer moves along the boundary of the square field of side 10 m in 40 seconds what will be the magnitude of the at the end of 2 metre into second

Answers

Answered by Anonymous

Correct Question

A farmer moves along the boundary of the square field of side 10 m in 40 seconds what will be the magnitude of the at the end of 2 metre in 20 second.

Solution

A farmer moves along the boundary of a square field of side 10 m in 40 sec.

Side of square = 10 m and time = 40 sec

Perimeter of square = 4 × side

= 4 × 10 = 40 m

We have to find the displacement of the farmer at the end of 2 min 20 sec.

Time = 2 min 20 sec

1 min = 60 sec

2 min = 2(60) = 120 sec

= 120 sec + 20 sec = 140 sec

Now,

In 1 sec distance covered by farmer = 40/40 = 1 m

So, in 140 sec distance covered by farmer = 1 × 140 = 140 m

Number of rotations to cover 140 m along the boundary = Distance/Perimeter

= 140/40 = 3.5 rounds

Therefore, the farmer takes 3.5 revolutions.

Let us assume that farmer is at the point A from the origin of the square field.

Now,

Displacement = diagonal of square

And from above we have a side of square = 10 m

So, displacement = 10√2 m

ғɪɴɴвαłσℜ: I'm I love with your answer

ғɪɴɴвαłσℜ: its awesome :O

Anonymous: :o thank you

Anonymous: Perfect ✌

Anonymous: thank you

Saby123: Nice Answer ❤❤

Answered by kailashmeena123rm

$\large{ \red{ \textsf{Correct question}}} \\$

a farmer moves along the boundary of the square field of side 10 m in 40 seconds what will be the magnitude of the displacement of farmer at the end of 2 minute 20 seconds from his initial position.

$\\ \large{\red{ \textsf{Answer}}} \\$

$\blue{figure}$

$<!DOCTYPE html><html><body><svg width="210" height="180"> <rect x="50" y="20" width="150" height="150" style="fill:none;stroke:black;stroke-width:5;fill-opacity:0.1;stroke-opacity:0.9" /> Sorry, your browser does not support inline SVG. </svg> </body></html>$

$\\$

$\blue{given}$

side = 10 m
he complete one round in 40 sec

$\blue{solution}$

$\rightarrow$ we have to find displacement after 2 min 40 sec

or 140 seconds

$\\$

$\rightarrow$ as he complete one round in 40 sec

no of rounds covered in 140 seconds = 140/40=3.5

$\\$

$\rightarrow$ so motion is this

he cover 3 rounds and a half round.He start from a point or corner of square and at end he will be at a opposite corner . since we have to find Magnitude .

$\\$

$\rightarrow$ Magnitude of displacement is distance between initial and final position

$\rightarrow$ use Pythagores theroem

$\\$

$\rightarrow$ displacement = $\sqrt{10^2 +10^2}$

$\rightarrow$ = 10 $\sqrt{2}$ m

Anonymous: osm ❤

ғɪɴɴвαłσℜ: really awesome

kailashmeena123rm: thanks❤

kailashmeena123rm: thanks ❤ MrMind

ғɪɴɴвαłσℜ: its ok dude no problem I'll do it always ʘ‿ʘ

Saby123: awesome rm

Previous Question

Next Question

a farmer moves along the boundary of the square field of side 10 m in 40 seconds what will be the magnitude of the at the end of 2 metre into second​

Answers

Correct Question

Solution

a farmer moves along the boundary of the square field of side 10 m in 40 seconds what will be the magnitude of the at the end of 2 metre into second