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Bas_tet [7]
3 years ago
7

As the earth rotates through one revolution, a person standing on the equator traces out a circular path whose radius is equal t

o the radius of the earth (6.38×10^6). What is the average speed of this person in meters per second? If we are standing in latitude 43° north what is our average speed?
Physics
1 answer:
mrs_skeptik [129]3 years ago
6 0

Answer:

The average speed will be "1038 mph".

Explanation:

Period for one revolution is:

T = 24 hours

  = 86,400 sec

x = c = 2πr

The given values is:

r = 6.38×10⁶ m

Now,

⇒  T=\frac{2 \pi r}{v}

Or,

⇒  v=\frac{2 \pi r}{T}

On substituting the values, we get

       =\frac{2 \pi (6.38\times 10^6)}{86,400}

       =464 \ min/sec

       =(464 \frac{min}{sec}) (\frac{3600 \ sec}{1 \ hour} )(\frac{1 \ mi}{1609 \ m} )

       =1038 \ mph

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Explanation:

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2 years ago
A diffraction grating is illuminated simultaneously with red light of wavelength 670 nm and light of an unknown wavelength. The
marusya05 [52]

Answer:

dsin∅ = m× λ

so, dsin∅red = 3(670nm)

also, dsin∅? =5λ?

however ,if they overlap then dsin∅red = dsin∅?

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2 years ago
If a soap bubble is 115 nm thick, what wavelength is most strongly reflected at the center of the outer surface when illuminated
sp2606 [1]

Answer:

611.8 nm

Explanation:

n_{oil} = Index of refraction of soap bubble  = 1.33

t = thickness of the soap bubble = 115 nm = 115 x 10⁻⁹ m

\lambda = wavelength of light = ?

m = order = 0

For reflection , the necessary condition is

2 n_{oil} t = (m + 0.5) \lambda

2 (1.33)(115\times 10^{-9})= (0 + 0.5) \lambda

\lambda = 6.118\times 10^{-7}

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7 0
2 years ago
An electron that has a velocity with x component 2.6 × 106 m/s and y component 3.2 × 106 m/s moves through a uniform magnetic fi
elena55 [62]

Answer:

a) \vec F_{B} = 8.766\times 10^{-14}\,T\,k, b) \vec F_{B} = -8.766\times 10^{-14}\,T\,k

Explanation:

a) The magnetic force experimented by a particle has the following vectorial form:

\vec F_{B} = q\cdot \vec v \times \vec B

The charge of the electron is equal to -1.602\times 10^{-19}\,C. Then, cross product can be solved by using determinants:

\vec F_{B} = \begin{vmatrix}i&j&k\\-4.165\times 10^{-13}\, C\cdot \frac{m}{s} &-5.126\times 10^{-13}\,C\cdot \frac{m}{s} &0\,C\cdot \frac{m}{s} \\0.041\,T&-0.16\,T&0\,T\end{vmatrix}

The magnetic force is:

\vec F_{B} = 8.766\times 10^{-14}\,T\,k

b) The charge of the proton is equal to 1.602\times 10^{-19}\,C. Then, cross product has the following determinant:

\vec F_{B} = \begin{vmatrix}i&j&k\\4.165\times 10^{-13}\, C\cdot \frac{m}{s} &5.126\times 10^{-13}\,C\cdot \frac{m}{s} &0\,C\cdot \frac{m}{s} \\0.041\,T&-0.16\,T&0\,T\end{vmatrix}

The magnetic force is:

\vec F_{B} = -8.766\times 10^{-14}\,T\,k

8 0
2 years ago
A ball is thrown up onto a roof, landing 4 sec later at height of 20m above the release level. The balls path just before landin
Kobotan [32]
<span>c) what is the angle (relative to the horizontal) of the balls initial velocity? </span>
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3 years ago
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