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Bond [772]
3 years ago
13

An airplane pilot sets a compass course due west and maintains an airspeed of 210 km/h. After flying for a time of 0.490 h, she

finds herself over a town a distance 121 km west and a distance 20 km south of her starting point.Find the magnitude of the wind velocity.
Physics
2 answers:
tankabanditka [31]3 years ago
5 0

Answer:

V_{A/E}=55.23km/h\\or\\V_{A/E}=15.342m/s

Explanation:

The velocity of plane relative to earth is given by:

V_{P/E}=\frac{d}{t}\\V_{P/E}=\frac{[-121i-20j]km}{0.490h}  \\V_{P/E}=[-247i-41j]km/h

As the from given data.The velocity of plane relative to air is:

V_{P/A}=[-210i]km/h

According to relative motion of velocity of the air relative to earth given by:

V_{A/E}=V_{P/E}-V_{P/A} \\V_{A/E}=[(-247i-41j)km/h]-(-210i)km/h\\V_{A/E}=[-37i-41j]km/h\\

The magnitude is given as:

V_{A/E}=\sqrt{(37)^{2}+(41)^{2}  }\\ V_{A/E}=55.23km/h\\or\\V_{A/E}=15.342m/s

Andrei [34K]3 years ago
5 0

Answer:

55km/h , 47.87° south west or 227.87°

Explanation:

If there is no wind then the plane would be at a distance

d=(210 km/h)(0.49h) = 102.9 km west of the starting point.

if the wind is blown then then the plane is 121 km - 102.9 km = 18.1 km west and 20 km south in the time 0.49 h.

Velocity of the plane in west direction=(18.1 km) / 0.49 h = 36.9 km/h

velocity of the plane in south direction=(20 km) / 0.49 h = 40.8 km/h

Now the wind velocity is

\sqrt{(36.9^2 + 40.8^2}

= 55.01 km/h

The direction is

θ =tan⁻¹ (40.8 / 36.9)

θ = 47.87° south west or 227.87°

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A 50g ball is released from rest 1.0 above the bottom of thetrack
ludmilkaskok [199]

Answer:

The maximum height of the ball is 2 m.

Explanation:

Given that,

Mass of ball = 50 g

Height = 1.0 m

Angle = 30°

The equation is

y=\dfrac{1}{4}x^2

We need to calculate the velocity

Using conservation of energy

\Delta U_{i}+\Delta K_{i}=\Delta K_{f}+\Delta U_{f}

Here, ball at rest so initial kinetic energy is zero and at the bottom the potential energy is zero

\Delta U_{i}=\Delta K_{f}

Put the value into the formula

mgh=\dfrac{1}{2}mv^2

Put the value into the formula

50\times10^{-3}\times9.8\times1.0=\dfrac{1}{2}\times50\times10^{-3}\times v^2

v^2=\dfrac{2\times50\times10^{-3}\times9.8\times1.0}{50\times10^{-3}}

v=\sqrt{19.6}

v=4.42\ m/s

We need to calculate the maximum height of the ball

Using again conservation of energy

\dfrac{1}{2}mv^2=mgh

Here, h = y highest point

Put the value into the formula

\dfrac{1}{2}\times50\times10^{-3}\times(4.42)^2=50\times10^{-3}\times9.8\times h

y=\dfrac{0.5\times(4.42)^2}{9.8}

y=0.996\ m

Put the value of y in the given equation

y=\dfrac{1}{4}x^2

x^2=4\times0.996

x=\sqrt{4\times0.996}

x=1.99\ m\ \approx 2 m

Hence, The maximum height of the ball is 2 m.

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Answer: The option (C) is correct. The energy travels horizontally.

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The distribution of the energy among the particles starts from center and it will carry out outward. This disturbance will occur horizontally.

Therefore, the energy travels horizontally.

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adoni [48]

Answer:

a) It will take you 16.94 min to climb the stairs.

b) The numbers of flights are 123 flights.

Explanation:

a) Let the energy consumption while climbing the stairs be 685 W in kcal/min is given by:

rate of energy consumption = (685 W)×[(0.01433kcal/min)/1 W]

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the time it takes to climb the stairs is given by:

t = E/P

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 = 16.94 min

Therefore, it will take you 16.94 min to climb the stairs.

b) Given t is the time it takes to climb stairs and N is the number of stairs to climb per min, then the number of stairs is:

n = t×N

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given the number of stairs each flight as N1, then the number of flights is given by:

flights = n/N1

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