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saul85 [17]
2 years ago
11

Looking down on a Northern Hemisphere extratropical cyclone, surface winds blow ________ about the center. a.counterclockwise an

d inward b.clockwise and outward
Physics
1 answer:
Lesechka [4]2 years ago
4 0

Answer:

counterclockwise and inward

Explanation:

Extratropical cyclones or mid-latitude cyclones are the low-pressure areas, along with anticyclones of the high-pressure areas that drive weather over much of the Earth.

<u>Due to the Coriolis effect, the flow of the wind around the extratropical cyclone is counterclockwise in direction in the northern hemisphere, and clockwise in direction in the southern hemisphere.</u>

<u>It forms a comma shape which drives the forces inward the center.</u>

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Compare and contrast how sound waves behave when they
Ghella [55]

Answer:

As sound waves move (or more accurately, when they travel by transferring their energy) they interact with physical objects. Soft surfaces will absorb sound while hard surfaces will reflect it. .Hard surfaces reflect sound and soft surfaces absorb sound.

The convex mirror has a reflecting surface that curves outward, resembling a portion of the exterior of a sphere. Light rays parallel to the optical axis are reflected from the surface in a direction that diverges from the focal point, which is behind the mirror

6 0
2 years ago
A photon of wavelength 7.33 pm scatters at an angle of 157° from an initially stationary, unbound electron. What is the de Brogl
Ann [662]

Answer:

4.63 p.m.

Explanation:

The problem given here can be solved by the Compton effect which is expressed as

\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)

here, \lambda  is the initial photon wavelength, \lambda^{'} is the scattered photon wavelength, h is he Planck's constant, m_e is the free electron mass, c is the velocity of light, \theta  is the angle of scattering.

Given that, the scattering angle is, \theta=157^{\circ}

Putting the respective values, we get

\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8}  } (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42(1-cos157^\circ ) p.m.

Therfore,

\lambda^{'}-\lambda=4.64 p.m.

Here, the photon's incident wavelength is \lamda=7.33pm

So,

\lambda^{'}=7.33+4.64=11.97 p.m

From the conservation of momentum,

\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}

here, \vec{P_\lambda} is the initial photon momentum, \vec{P_{\lambda^{'}}} is the final photon momentum and \vec{P_e} is the scattered electron momentum.

Expanding the vector sum, we get

P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta

Now expressing the momentum in terms of De-Broglie wavelength

P=h/\lambda and putting it in the above equation we get,

\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}

Therfore,

\lambda_{e}=\frac{7.33\times 11.97}{\sqrt{7.33^{2}+11.97^{2}-2\times 7.33\times 11.97\times cos157^\circ }} p.m.\\\lambda_{e}=\frac{87.7401}{18.935} = 4.63 p.m.

This is the de Broglie wavelength of the electron after scattering.

8 0
3 years ago
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac
Y_Kistochka [10]

Answer:

See Explanation

Explanation:

a) We know that;

v = λf

Where;

λ = wavelength of the wave

f = frequency of the wave

v = velocity of the wave

So;

T = 2 * 2.10 s = 4.2 s

Hence f = 1/4.2 s

f = 0.24 Hz

The wavelength =  6.5 m

Hence;

v = 6.5 m * 0.24 Hz

v = 1.56 m/s

b)The amplitude of the wave is;

A =  0.600 m/2 = 0.300 m

c) Since the wave speed does not depend on the amplitude of the wave then the answer in (a) above remains the same

Where d = 0.30 m

A = 0.30 m/2 = 0.15 m

6 0
3 years ago
ITS TIMED PLEASE HELP !!!!!
dusya [7]

Answer: direct linear

Explanation:

3 0
3 years ago
HELP PLEASE 20 POINTS SHOW WORK, ALL EQUATIONS
nataly862011 [7]

Answer:

s = 3 m

Explanation:

Let t be the time the accelerating car starts.

Let's assume the vehicles are point masses so that "passing" takes no time.

the position of the constant velocity and accelerating vehicles are

s = vt = 40(t + 2)  cm

s = ½at² = ½(20)(t)² cm

they pass when their distance is the same

½(20)(t)² = 40(t + 2)

10t² = 40t + 80

0 = 10t² - 40t - 80

0 = t² - 4t - 8

t = (4±√(4² - 4(1)(-8))) / 2(1)

t = (4± 6.928) / 2  ignore the negative time as it has not occurred yet.

t = 5.464 s

s = 40(5.464 + 2) = 298.564 cm

300 cm when rounded to the single significant digit of the question numerals.

7 0
3 years ago
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