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Ira Lisetskai [31]
4 years ago
15

A plane takes off from an airport and flies to town A, located d1 = 335 km from the airport in the direction 20.0° north of east

. The plane then flies to town B, located d2 = 245 km at 30.0° west of north from town A. Use graphical methods to determine the distance and direction from town B to the airport. (Enter the distance in km and the direction in degrees south of west.)
Physics
1 answer:
gtnhenbr [62]4 years ago
4 0

Answer:

R = 545.38 m ; θ = 28.43°

Explanation:

given,

for town A : d₁ = 335 km at an direction of 20° north of east

for town B : d₂ = 245 km at  30.0° west of north from town A

x = 335 cos 20° + 245 sin 30°

x = 437.29 m

y = 335 sin 20° + 245 cos 30°

y = 326.75 m

resultant = \sqrt{x^2+y^2}

R =\sqrt{437.29^2+326.75^2}

R = 545.38 m

tan \theta = \dfrac{326.75}{437.29}

θ = 28.43°

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Explanation:

As the final and initial velocities are known it is possible then the kinetic energy is possible to calculate for each instant.

By definition, the kinetic energy is:

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Expressing the initial and final kinetic energy for cars A and B:

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Since the masses are equals:

m=ma=mb

For the known velocities, the kinetics energies result:

ki=0.5*mVa_{i}^2

ki=0.5*m(35 m/s)^2=612.5m^2/s^2*m

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kf=0.5*m(25 m/s)^2=312.5m^2/s^2*m

The lost energy in the collision is the difference between the initial and final kinectic energies:

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Finally the relation between the lost and the initial kinetic energy:

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4 years ago
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Answer:

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Equate 1 and 2 to have,

mg = \frac{GMm}{r^{2} }

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