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Korvikt [17]
2 years ago
5

An airplane flies between two points on the ground that are 500 km apart. The destination is directly north of the origination o

f the flight. The plane flies with an air speed of 120 m/s. If a constant wind blows at 10.0 m/s due west during the flight, what direction must the plane fly relative to north to arrive at the destination? Consider: east to the right, west to the left, north upwards and south downwards
Physics
1 answer:
Ann [662]2 years ago
5 0

Answer:

   θ = 4.78º

with respect to the vertical or 4.78 to the east - north

Explanation:

This is a velocity compound exercise since it is a vector quantity.

The plane takes a direction, the air blows to the west and the result must be to the north, let's use the Pythagorean theorem to find the speed

                  v_fly² = v_nort² + v_air²

                  v_nort² = v_fly² + - v_air²

Let's use trigonometry to find the direction of the plane

        sin θ = v_air / v_fly

        θ = sin⁻¹ (v_air / v_fly)

         

let's calculate

        θ = sin⁻¹ (10/120)

         θ = 4.78º

with respect to the vertical or 4.78 to the north-east

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The answer is C. You must divide your wavelength and your frequency to get your answer.
8 0
3 years ago
Calculate the energy, wavelength, and frequency of the emitted photon when an electron moves from an energy level of -3.40 eV to
jasenka [17]

Answer:

(a) The energy of the photon is 1.632 x 10^{-8} J.

(b) The wavelength of the photon is 1.2 x 10^{-17} m.

(c) The frequency of the photon is 2.47 x 10^{25} Hz.

Explanation:

Let;

E_{1} = -13.60 ev

E_{2} = -3.40 ev

(a) Energy of the emitted photon can be determined as;

E_{2} - E_{1} = -3.40 - (-13.60)

           = -3.40 + 13.60

           = 10.20 eV

           = 10.20(1.6 x 10^{-9})

E_{2} - E_{1} = 1.632 x 10^{-8} Joules

The energy of the emitted photon is 10.20 eV (or 1.632 x 10^{-8} Joules).

(b) The wavelength, λ, can be determined as;

E = (hc)/ λ

where: E is the energy of the photon, h is the Planck's constant (6.6 x 10^{-34} Js), c is the speed of light (3 x 10^{8} m/s) and λ is the wavelength.

10.20(1.6 x 10^{-9}) = (6.6 x 10^{-34} * 3 x 10^{8})/ λ

λ = \frac{1.98*10^{-25} }{1.632*10^{-8} }

  = 1.213 x 10^{-17}

Wavelength of the photon is 1.2 x 10^{-17} m.

(c) The frequency can be determined by;

E = hf

where f is the frequency of the photon.

1.632 x 10^{-8}  = 6.6 x 10^{-34} x f

f = \frac{1.632*10^{-8} }{6.6*10^{-34} }

 = 2.47 x 10^{25} Hz

Frequency of the emitted photon is 2.47 x 10^{25} Hz.

6 0
2 years ago
One gallon of gasoline in an automobiles engine produces on average 9.50 kg of carbon dioxide, which is a greenhouse gas; that i
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The annual production of carbon dioxide is 124121.49×10^{6}[/tex] kg.

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Fuel consumed=326.63*40*10^6=13065.42 million gallon,

Now we can calculate the annual production of carbon dioxide in the USA

CO2 production rate=9.50*13065.42=124121.49*10^6 kg

Therefore the annual production of carbon dioxide in USA is 124121.49×10^{6}[/tex] kg

4 0
3 years ago
A 35 kg child goes down a playground that is inclined at an angle of 27.5 degrees above the horizontal. Find the acceleration if
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Answer:

4.16m/s²

Explanation:

According to Newtons second law;

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Fm is the moving force

\mu is  the coefficient of kinetic friction between the child and the slide

m is the mass

g is the acceleration due to gravity

a is the acceleration of the child

Substitute the given values and get the acceleration as shown;

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158.38-12.88 = 35a

145.49 = 35a

a = 145.49/35

a = 4.16m/s²

Hence the acceleration of the body is  4.16m/s²

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