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m_a_m_a [10]
3 years ago
7

. Solve the following problem using analytical techniques: Suppose you walk 18.0 m straight west and then 25.0 m straight north.

How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position?
Physics
1 answer:
KiRa [710]3 years ago
7 0

Answer:

H = 30.8 m

\theta =54.25^0

Explanation:

given,

first walks straight west = 18 m

then straight in north = 25 m

the distance from the start point

H² = B² + P²

H² = 18² + 25²

H = \sqrt{18^2+25^2}

H = \sqrt{324+625}

H = 30.8 m

the angle between

tan \theta = \dfrac{P}{B}

tan \theta = \dfrac{25}{18}

\theta = tan^{-1}(\dfrac{25}{18})

\theta =54.25^0

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Answer:

0.03167 m

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Explanation:

x = Compression of net

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g = Acceleration due to gravity = 9.81 m/s²

The potential energy and the kinetic energy of the system is conserved

P_i=P_f+K_s\\\Rightarrow mgh_i=-mgx+\frac{1}{2}kx^2\\\Rightarrow k=2mg\frac{h_i+x}{x^2}\\\Rightarrow k=2\times 65\times 9.81\frac{18+1.1}{1.1^2}\\\Rightarrow k=20130.76\ N/m

The spring constant of the net is 20130.76 N

From Hooke's Law

F=kx\\\Rightarrow x=\frac{F}{k}\\\Rightarrow x=\frac{65\times 9.81}{20130.76}\\\Rightarrow x=0.03167\ m

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If h = 35 m

From energy conservation

65\times 9.81\times (35+x)=\frac{1}{2}20130.76x^2\\\Rightarrow 10065.38x^2=637.65(35+x)\\\Rightarrow 35+x=15.785x^2\\\Rightarrow 15.785x^2-x-35=0\\\Rightarrow x^2-\frac{200x}{3157}-\frac{1000}{451}=0

Solving the above equation we get

x=\frac{-\left(-\frac{200}{3157}\right)+\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}, \frac{-\left(-\frac{200}{3157}\right)-\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}\\\Rightarrow x=1.52, -1.45

The compression of the net is 1.52 m

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