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3 years ago

A frog hops 5m east and 2m north. What is the magnitude of the frogs total displacement in m?

Physics

1 answer:

TiliK225 [7]3 years ago

7 0

As per the question a frog jumps 5 m towards east.

Frog again jumps 2 m north.

Let the displacement along east is denoted by vector A and the displacement towards north is denoted as vector B.

Hence magnitude of A = 5 m

Magnitude of B = 2 m

We are asked to calculate the total displacement.

Here the angle between them is 90 degree as A is towards east and B is towards north.

As per parallelogram law of vector addition,the magnitude of total displacement [R] will be-

$R=\sqrt{ A^{2} +B^{2}+2AB\cos\theta}$

$=\sqrt{ 5^{2}+ 2^{2}+2*5*2 cos 90}$

$=\sqrt{25+4+0} m$ [cos90= 0]

$=\sqrt{29} m$

$= 5.38516 m$ [ans]

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3 years ago

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3 years ago

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Answer:

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Explanation:

Given that,

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3 years ago

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Answer:

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