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Natasha_Volkova [10]
3 years ago
6

A monkey running through the jungle goes 0.198 km straight to the East, then turns 15.8° deflection from straight East toward th

e South, and then goes 145 m in a straight line in the diagonal direction he is now facing. What is the displacement of the monkey? (put your answer in standard units)
Physics
1 answer:
inn [45]3 years ago
5 0

Answer:

|\vec r|=339.82\ m

\theta=-6.67^o

Explanation:

<u>Displacement </u>

It's a vector magnitude that measures the space traveled by a particle between an initial and a final position. The total displacement can be obtained by adding the vectors of each individual displacement. In the case of two displacements:

\vec r=\vec r_1+\vec r_2

Given a vector as its polar coordinates (r,\theta), the corresponding rectangular coordinates are computed with

x=rcos\theta

y=rsin\theta

And the vector is expressed as

\vec z==

The monkey first makes a displacement given by (0.198 km,0°). The angle is 0 because it goes to the East, the zero-reference for angles. Thus the first displacement is

\vec r_1==\ km=\ m

The second move is (145 m , -15.8°). The angle is negative because it points South of East. The second displacement is

\vec r_2==\ m

The total displacement is

\vec r=\ m+\ m

\vec r=\ m

In (magnitude,angle) form:

|\vec r|=\sqrt{337.52^2+(-39.48)^2}=339.82\ m

\boxed{|\vec r|=339.82\ m}

\displaystyle tan\theta=\frac{-39.48}{337.52}=-0.1169

\boxed{\theta=-6.67^o}

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Question 1.

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now, we know :

=》

p.e =  mgh

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63000 =  \dfrac{1}{2}  \times 4500 \times  {v}^{2}

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Answer:

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