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valina [46]
2 years ago
13

Bryan Allen pedaled a human-powered aircraft across the English channel from the cliffs of Dover to Cap Gris-Nez on June 12, 197

9. He flew for 169 minutes at an average velocity of 3.53 m/s in a direction 45.0∘ degrees south of east. Allen encountered a headwind averaging 2.00 m/s almost precisely in the opposite direction of his motion relative to the earth. What was his average velocity relative to the air?
Physics
1 answer:
tigry1 [53]2 years ago
4 0

Answer:

The answer is "5.53 \ \frac{m}{s}"

Explanation:

apply the formula for calculating the average velocity to the relative air

V_{PG} =V_{PA}+V_{AG}

\Rightarrow  V_{PA} = V_{PG} -V_{AG}

Given value:

V_{AG} = -2 \ \frac{m}{s}

V_{PG} =3.53

\Rightarrow  V_{PA} = 3.53 - (-2) \\\\\Rightarrow  V_{PA} = 3.53 +2 \\\\\Rightarrow  V_{PA} = 5.53  \\\\

The final answer is "5.53 \ \frac{m}{s}" in the south-east direction.

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3 years ago
A 0.6 kg block attached to a spring of force constant 13.6 N/m oscillates with an amplitude of 9 cm. Find the maximum speed of t
mash [69]

Answer:

1) 0.43 meters per second

2) 0.21 meters per second

3) 1.02 \frac{m}{s^{2}}

4) 0.66 seconds

Explanation:

part 1

By conservation of energy, the maximum kinetic energy (K) of the block is at equilibrium point where the potential energy is zero. So, at the equilibrium kinetic energy is equal to maximum potential energy (U):

K=U

\frac{mv^2}{2}=\frac{kx_{max}^2}{2}

With m the mass, v the speed, k the spring constant and xmax the maximum position respect equilibrium position. Solving for v

v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.09m)^2}{0.6}}=0.43\frac{m}{s}

part 2

Again by conservation of energy we have kinetic energy equal potential energy:

\frac{mv^2}{2}=\frac{kx_{max}^2}{2}=

v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.045m)^2}{0.6}}=0.21\frac{m}{s}

part 3

Acceleration can be find using Newton's second law:

F=ma

with F the force, m the mass and a the acceleration, but elastic force is -kx, so:

-kx=ma

a= -\frac{kx}{m}=-\frac{(13.6)(0.045)}{0.6}=-1.02\frac{m}{s^{2}}

part 4

The period of an oscillator is the time it takes going from one extreme to the other one, that is going form 4.5 cm to -4.5 cm respect the equilibrium position. That period is:

T=2\pi\sqrt{\frac{m}{k}}=T=2\pi\sqrt{\frac{0.6}{13.6}}=1.32s

So between 0 and 4.5 cm we have half a period:

t=\frac{T}{2}=0.66s

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