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3 years ago

Waves diffract the most when their wavelength is

Physics

1 answer:

djverab [1.8K]3 years ago

5 0

<h2>Answer: about the same size of the gap or slit</h2>

Diffraction happens when a wave (mechanical or electromagnetic wave, in fact, any wave) meets an obstacle or a slit .When this occurs, the wave bends around the corners of the obstacle or passes through the opening of the slit that acts as an obstacle, forming multiple patterns with the shape of the aperture of the slit.

Note that the principal condition for the occurrence of this phenomena is that the obstacle must be comparable in size (similar size) to the size of the wavelength.

In other words, when the gap (or slit) size is larger than the wavelength, the wave passes through the gap and does not spread out much on the other side, but when the gap size is equal to the wavelength, maximum diffraction occurs.

Therefore:

<h2>Waves diffract the most when their wavelength is <u>about the same size of the gap </u></h2>

<u />

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.. As you increase the

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Answer:

air pressure increases and temperature decreases

Explanation:

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3 years ago

Which forces are acting on the student and the skateboard in the instant in which they are pushing off the wall? (Select all tha

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E all of the answers above correlate to the student and his skateboard

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2 years ago

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A ball is dropped from a height of 20 meters. At what helght does the ball have a velocity of 10 meters/second?

Sunny_sXe [5.5K]

Answer:

5.10 meters.

Explanation:

v²=u²+2gh

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or, 19.6h=100

or, h=5.10 meters

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3 years ago

Suppose you pour 0.250 kg of 20.0°C water into a 0.600 kg aluminum pan off the stove with a temperature of 173°C. Assume that th

lapo4ka [179]

Answer:

$T_f=5.0116^{\circ}C$

Explanation:

Given:

mass of water, $m_w=0.25\ kg$

initial temperature of water, $T_i_w=20^{\circ}C$

initial temperature of pan, $T_i_p=173^{\circ}C$

mass of pan, $m_p=0.6\ kg$

mass of water evapourated, $m_v=0.03\ kg$

specific heat of water, $c_w=4186\ J.kg^{-1}.K^{-1}$

specific heat of aluminium pan, $c_a=900\ J.kg^{-1}.K^{-1}$

latent heat of vapourization, $L=2256000\ J.kg^{-1}$

<u>Using the equation of heat:</u>

<em>Here, initially certain mass of water is vapourised first and then the remaining mass of water comes in thermal equilibrium with the pan.</em>

$m_p.c_a.(T_{ip}-T_f)=m_v.L+(m_w-m_v).c_w.(T_f-T_{iw})$

$0.6\times 900\times (173-T_f)=0.03\times 2256000+(0.25-0.03)\times 4186\times (T_f-20)$

$T_f=5.0116^{\circ}C$

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3 years ago

Pikes Peak near Denver, Colorado, has an elevation of 14,110 ft. Calculate the pressure at this elevation using three different

kramer

Answer:

a) P = 1240 lb/ft^2

b) P = 1040 lb/ft^2

c) P = 1270 lb/ft^2

Explanation:

Given:

- P_a = 2216.2 lb/ft^2

- β = 0.00357 R/ft

- g = 32.174 ft/s^2

- T_a = 518.7 R

- R = 1716 ft-lb / slug-R

- γ = 0.07647 lb/ft^3

- h = 14,110 ft

Find:

(a) Determine the pressure at this elevation using the standard atmosphere equation.

(b) Determine the pressure assuming the air has a constant specific weight of 0.07647 lb/ft3.

Solution:

- The standard atmospheric equation is expressed as:

P = P_a* ( 1 - βh/T_a)^(g / R*β)

(g / R*β) = 32.174 / 1716*0.0035 = 5.252

P = 2116.2*(1 - 0.0035*14,110/518.7)^5.252

P = 1240 lb/ft^2

- The air density method which is expressed as:

P = P_a - γ*h

P = 2116.2 - 0.07647*14,110

P = 1040 lb/ft^2

- Using constant temperature ideal gas approximation:

P = P_a* e^ ( -g*h / R*T_a )

P = 2116.2* e^ ( -32.174*14110 / 1716*518.7 )

P = 1270 lb/ft^2

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3 years ago

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