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SSSSS [86.1K]
3 years ago
12

Canadian geese migrate essentially along a north-south direction for well over a thousand kilometers in some cases, traveling at

speeds up to about 100 km/h. If one such bird is flying at 100 km/h relative to the air, but there is a 34.0 km/h wind blowing from west to east.At what angle relative to the north-south direction should this bird head so that it will be traveling directly southward relative to the ground?
Physics
1 answer:
Wewaii [24]3 years ago
5 0

Answer:

θ=19.877⁰

Explanation:

Given data

Velocity Va=34.0 km/h

Velocity Va=100 km/h

To find

Angle θ

Solution  

We want the bird to fly with velocity Vb=100 km/h with an angle θ relative to the ground so that the bird fly due south relative to the ground.From figure which is attached we got

Sinθ=(Va/Vb)

Sinθ=(34.0/100)

θ=Sin⁻¹(34.0/100)

θ=19.877⁰

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Ne W2
levacccp [35]

Answer:

3.82746e+26 watts

Explanation:

There are two ways to solve this problem. One way is to use the equation

L = 4πσR²T⁴

where

L = the sun's bolometric (all-spectrum) luminous power

σ = 5.670374419e-8 W m⁻² K⁻⁴ = the Stefan-Boltzmann constant

R = 6.957e+8 meters = the sun's radius

T = 5771.8 K = the sun's effective temperature

You find that

L = 3.82746e+26 watts

The other way to solve the problem is to use the Planck integral for radiant flux.

L = 4π²R ∫(v₁,v₂) 2hv³/{c² exp[hv/(kT)]−1} dv

where

h = 6.62607015e-34 J sec

c = 299792458 m sec⁻¹

k = 1.380649e-23 J K⁻¹

v₁ = 0 = frequency band lower bound, in Hz

v₂ = ∞ = frequency band upper bound, in Hz

You find, once again, that

L = 3.82746e+26 watts

The advantage of using the Planck integral becomes clear when you want to calculate the sun's luminous power only in a specific band, rather than across the entire spectrum. For example, if we do the calculation again, except that we use

v₁ = 4.1e+14 = frequency band lower bound, in Hz

v₂ = 7.7e+14 Hz = frequency band upper bound, in Hz

restricting ourselves to the visible spectrum. We find that

L (visible) = 1.56799e+26 watts

So the fraction of the sun's luminosity that is in the visible spectrum is

L (visible) / L = 0.4096686

5 0
4 years ago
A fisherman sitting on the end of a pier notices that 6 wave crests pass him in 3 seconds. What is the frequency of the waves?.
dalvyx [7]

Answer:

2 Hz

Explanation:

f = frequency

n = wave crest

t = time

f = n/t → f = 6/3 = 2

4 0
3 years ago
Several scientists from different countries are asked to examine the results of an experiment before a journal will print it. Wh
DENIUS [597]
The term that is being described by the definition given above is called PEER REVIEW. What is done in the peer review is that, the experiment is being evaluated or checked first by people who are also working on the same field. So the answer for this is option C.
8 0
4 years ago
Chapter 38, Problem 001
Norma-Jean [14]

Answer:

a) \lambda=2.95x10^{-6}m

b) infrared region

Explanation:

Photon energy is the "energy carried by a single photon. This amount of energy is directly proportional to the photon's electromagnetic frequency and is inversely proportional to the wavelength. If we have higher the photon's frequency then we have higher its energy. Equivalently, with longer the photon's wavelength, we have lower energy".

Part a

Is provide that the smallest amount of energy that is needed to dissociate a molecule of a material on this case 0.42eV. We know that the energy of the photon is equal to:

E=hf

Where h is the Planck's Constant. By the other hand the know that c=f\lambda and if we solve for f we have:

f=\frac{c}{\lambda}

If we replace the last equation into the E formula we got:

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

And if we solve for \lambda we got:

\lambda =\frac{hc}{E}

Using the value of the constant h=4.136x10^{-15} eVs we have this:

\lambda=\frac{4.136x10^{15}eVs (3x10^8 \frac{m}{s})}{0.42eV}=2.95x10^{-6}m

\lambda=2.95x10^{-6}m

Part b

If we see the figure attached, with the red arrow, the value for the wavelenght obtained from part a) is on the infrared region, since is in the order of 10^{-6}m

4 0
3 years ago
You are dragging a block on a surface with friction at a steady speed of 2 m/s and exert a force of 5 N to do so. What is the fo
m_a_m_a [10]

Answer:

Friction = 5 N

Explanation:

As we know that block is moving at constant speed

So the acceleration of the block is zero

So we will have

F_{net} = 0

for net force to be zero

Force exerted on the object by external system must be counter balanced by the force of friction

So we have

F_{ex} = F_f

so we have

F_f = 5 N

6 0
4 years ago
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