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kumpel [21]
3 years ago
13

A pilot must fly his plane due north to reach his destination. The plane can fly at 300 km/h in still air. A wind is blowing out

of the northeast at 90 km/h.
(a) What is the speed of the plane relative to the ground?

(b) In what direction must the pilot head her plane to fly due north?
Physics
1 answer:
ikadub [295]3 years ago
4 0

Answer:

Answer explained below

Explanation:

The situation in this problem suggests that the pilot wants to fly the plane towards north with a velocity of magnitude 300km/hr. But the wind is flowing from Northeast with the magnitude of 90km/hr.

(a) so there's a wind resistance in action, which will prevent the plane from reaching its desired speed of 300km/hr. That means the velocity of the plane with respect to ground will be less than 300km

The plane needs to have a northward heading but the wind is forcing it towards southwest. So the pilot has to compensate for this resistance by changing the direction of the plane i.e. flying towards northeast so that the effect of the wind is nullified and plane keeps flying north at the expense of velocity . It's a simple matter of vector addition by triangle law to find out the resultant velocity as the author has explained. So the theory used by the author is correct and the math checks out to.

(b) the destination of the plane is due north. So it has to fly towards north with respect to ground. So it's velocity vector must be pointing north with respect to the ground. But the wind velocity vector is oriented from Northeast - northwest direction. i e. tail of the wind velocity vector is towards northeast and the head is towards northwest and attached to the head of the northward velocity of the plane. This northward velocity is the resultant velocity maintained by the plane by flying towards northeast at a velocity of 300km/hr. That gives us the resultant northward velocity by triangle law of vector addition to be 230km/hr.

Since we are dealing with a triangle of vectors,we can simply apply basic trigonometry to decide in what direction must the plane fly in order to maintain a northward heading. The author has used the correct relation to calculate the direction of the plane. To have a north heading,it means that the wind resistance is compensated by sacrificing some of the velocity.

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Answer:

The value is E_i  =  1.5596 *10^{-18} \  J

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From the question we are told that

The wavelength is \lambda  =  48.2 nm  =  48.2  *10^{- 9 }\  m

The velocity is v = 2.371*10^6 \ m/s

The mass of electron is m_e  =  9.109*10^{-31} \  kg

Generally the energy of the incident light is mathematically represented as

E =  \frac{h *  c}{\lambda}

Here c is the speed of light with value c =  3.0 *10^{8} \  m/s

h is the Planck constant with value h = 6.62607015 *  10^{-34 }  J\cdot s

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Generally the kinetic energy is mathematically represented as

E_k  =  \frac{1}{2} *  m_e * v^2

=> E_k  =  \frac{1}{2} *  9.109*10^{-31} * (2.371*10^6 )^2

=> E_k  =  2.56 *0^{-18} \  J

Generally the ionization energy is mathematically represented as

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Answer:

ω = 12.023 rad/s

α = 222.61 rad/s²

Explanation:

We are given;

ω0 = 2.37 rad/s, t = 0 sec

ω =?, t = 0.22 sec

α =?

θ = 57°

From formulas,

Tangential acceleration; a_t = rα

Normal acceleration; a_n = rω²

tan θ = a_t/a_n

Thus; tan θ = rα/rω² = α/ω²

tan θ = α/ω²

α = ω²tan θ

Now, α = dω/dt

So; dω/dt = ω²tan θ

Rearranging, we have;

dω/ω² = dt × tan θ

Integrating both sides, we have;

(ω, ω0)∫dω/ω² = (t, 0)∫dt × tan θ

This gives;

-1[(1/ω_o) - (1/ω)] = t(tan θ)

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ω = ω_o/(1 - (ω_o × t × tan θ))

While;

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Thus, plugging in the relevant values;

ω = 2.37/(1 - (2.37 × 0.22 × tan 57))

ω = 12.023 rad/s

Also;

α = (2.37² × tan 57)/(1 - (2.37 × 0.22 × tan 57))²

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