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Oksi-84 [34.3K]
3 years ago
7

An airplane pilot wishes to fly directly westward. According to the weather bureau, a wind of 75.0km/hour is blowing southward.

The speed of the plane relative to the air (called the air speed) as measured by instruments aboard the plane is 310km/hour . In which direction should the pilot head?
1. The wind is blowing southward at 75.0km/hour . In what frame of reference is this speed measured?

a. The speed is measured relative to the plane.
b. The speed is measured relative to the air.
c. The speed is measured relative to the ground.

2. An airplane pilot wishes to fly directly westward. According to the weather bureau, a wind of 75.0km/hour is blowing southward. The speed of the plane relative to the air (called the air speed) as measured by instruments aboard the plane is 310km/hour . In which direction should the pilot head?

3. In this particular problem, you can compare the speed of the plane relative to the ground (the ground speed) to known quantities to evaluate the reasonableness of your answer. Find the ground speed of the plane, vP/G.
Physics
1 answer:
qaws [65]3 years ago
7 0

Answer:

a)  correct answer is C , b) 14º  from the west to the north, c)   v_{1g} = 300.79 km / h

Explanation:

This is a relative speed exercise using the addition of speeds.

1) when it is not specified regarding what is being measured, the medicine is carried out with respect to the Z Earth, therefore the correct answer is C

2 and 3) In this case we must compose the speed using the Pythagorean Theorem.

     v_{1a}² = v_{1g}² + v_{ag}²

where v_{1a} is the speed of the airplane with respect to the air, v_{1g} airplane speed with respect to the Earth, v_{ag} air speed with respect to the Earth

in this case let's clear the speed of the airplane with respect to the Earth

  v_{1g} = √(v_{1a}² - v_{ag}²)

 v_{1g} = √ (310² - 75²)

 v_{1g} = 300.79 km / h

we find the direction of the airplane using trigonometry

   sin θ = v_{ag} / v_{1a}

   θ = sin⁻¹ (v_{ag} /v_{1a})

   θ = sin⁻¹ (75/310)

   θ= 14º

the pilot must direct the aircraft at an angle of 14º from the west to the north

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Answer:

Part a)

T = 0.81 s

Part b)

v_x = 3.33 m/s

Part c)

v_y = 3.91 m/s

Part d)

\theta = 49.55 degree

Part e)

T = 1.11 s

Explanation:

Part a)

initial vertical position = 1.02 m

maximum height = 1.80 m

\Delta y = 1.80 - 1.02

\Delta y = 0.78 m

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0 - v_y^2 = 2(-9.81)(0.78)

v_y = 3.91 m/s

time taken by it to reach this height

v_y = v_i + at

0 = 3.91 - 9.81 t_1

t_1 = 0.39 s

Now when it again touch the ground then its speed is given as

v_f^2 - v_y^2 = 2a \Delta y

v_f^2 - 0 = 2(9.81)(1.80 - 0.95)

v_y = 4.08 m/s

time taken by it to reach this height

4.08 = v_i + at

4.08 = 0 + 9.81 t_2

t_2 = 0.42 s

T = t_1 + t_2

T = 0.81 s

Part b)

Horizontal velocity

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v_x = \frac{2.70}{0.81}

v_x = 3.33 m/s

Part c)

vertical velocity is the intial y direction velocity

v_y = 3.91 m/s

Part d)

Take off angle is given as

tan\theta = \frac{3.91}{3.33}

\theta = 49.55 degree

Part e)

initial vertical position = 1.20 m

maximum height = 2.50 m

\Delta y = 2.50 - 1.20

\Delta y = 1.30 m

v_f^2 - v_y^2 = 2a \Delta y

0 - v_y^2 = 2(-9.81)(1.30)

v_y = 5.05 m/s

time taken by it to reach this height

v_y = v_i + at

0 = 5.05 - 9.81 t_1

t_1 = 0.51 s

Now when it again touch the ground then its speed is given as

v_f^2 - v_y^2 = 2a \Delta y

v_f^2 - 0 = 2(9.81)(2.50 - 0.72)

v_y = 5.9 m/s

time taken by it to reach this height

5.9 = v_i + at

5.9 = 0 + 9.81 t_2

t_2 = 0.60 s

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A researcher studying the nutritional value of a new candy places a 4.60 g sample of the candy inside a bomb calorimeter and com
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Answer:

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Heat gained by the bomb calorimeter = c×ΔT

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Heat lost by the new candy = heat gained by the bomb calorimeter = 86618 J = 20702.2 calories

4.60 g of the new candy lost this amount of calories by undergoing combustion,

The amount of calories per g = 20702.2 calories/4.6 g = 4500.5 calories per gram

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just olya [345]
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Answertrue

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The distance covered is 25.9 m.

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