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3 years ago

3. All of the following are parts of a longitudinal wave EXCEPT

Physics

2 answers:

erastova [34]3 years ago

6 0

The answer is Crest and all the others are the part of a longitudinal wave

Goryan [66]3 years ago

3 0

Crest. Hope this helpled

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A spy camera is said to be able to read the numbers on a car's license plate. If the numbers on the plate are 4.30 cm apart, and

Maslowich

Answer:

D = 2.38 m

Explanation:

This exercise is a diffraction problem where we must be able to separate the license plate numbers, so we must use a criterion to know when two light sources are separated, let's use the Rayleigh criterion, according to this criterion two light sources are separated if The maximum diffraction of a point coincides with the first minimum of the second point, so we can use the diffraction equation for a slit

a sin θ = m λ

Where the first minimum occurs for m = 1, as in these experiments the angle is very small, we can approximate the sine to the angle

θ = λ / a

Also when we use a circular aperture instead of slits, we must use polar coordinates, which introduce a numerical constant

θ = 1.22 λ / D

Where D is the circular tightness

Let's apply this equation to our case

D = 1.22 λ / θ

To calculate the angles let's use trigonometry

tan θ = y / x

θ = tan⁻¹ y / x

θ = tan⁻¹ (4.30 10⁻² / 140 10³)

θ = tan⁻¹ (3.07 10⁻⁷)

θ = 3.07 10⁻⁷ rad

Let's calculate

D = 1.22 600 10⁻⁹ / 3.07 10⁻⁷

D = 2.38 m

4 0

3 years ago

Read 2 more answers

A proton has been accelerated from rest through a potential difference of -1000 v . part a what is the proton's kinetic energy,

wlad13 [49]

We can apply the law of conservation of energy here. The total energy of the proton must remain constant, so the sum of the variation of electric potential energy and of kinetic energy of the proton must be zero:
$\Delta U + \Delta K=0$
which means
$\Delta K = - \Delta U$
The variation of electric potential energy is equal to the product between the charge of the proton (q=1eV) and the potential difference ( $\Delta V=-1000 V$ ):
$\Delta U = q \Delta V=(1 eV)(-1000 V)=-1000 eV$
Therefore, the kinetic energy gained by the proton is
$\Delta K = -(-1000 eV)=1000 eV$
<span>And since the initial kinetic energy of the proton was zero (it started from rest), then this 1000 eV corresponds to the final kinetic energy of the proton.</span>

4 0

3 years ago

Read 2 more answers

Robert has just bought a new model rocket, and is trying to measure its flight characteristics. The rocket engine package claims

777dan777 [17]

Answer:

The work done by the drag force is given by 29.96 J

Explanation:

Given :

Thrust force $F = 12.3$ N

Displacement $d = 10.2$ m

Mass of rocket $m = 0.663$ Kg

From work energy theorem,

$W = \Delta K$

$W_{t} - Wd - W_{g} = KE$

Where $W_{t} =$ thrust work $W_{g} =$ gravitational work

$KE = 12.3 \times 10.2 -Wd - 0.663 \times 9.8 \times 10.2$

$KE = 59.2 -Wd$

After cutoff kinetic energy is converted into potential energy,

$KE = Wd' + mg\Delta h$

Put value of KE

$59.2 -Wd = Wd' + 0.663 \times 9.8 \times 4.5$

Work done by drag force is given by,

$Wd'+Wd = 59.2 -29.23$

$= 29.96$ J

Therefore, the work done by the drag force is given by 29.96 J

5 0

3 years ago

How can two polarizing filters be used to show that light has some properties of a wave

DIA [1.3K]

Polarizing filters: used to show light has some properties of a wave in that way of property is that light can be thought of as traveling forward in waves, with the <span>wave. sorry if this is too late but google is good</span>

6 0

3 years ago

An insect 5.25 mm tall is placed 25.0 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the righ

Zigmanuir [339]

Answer:

(A) therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

(B) therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

Explanation:

height of the insect (h) = 5.25 mm = 0.525 cm

distance of the insect (s) = 25 cm

radius of curvature of the flat left surface (R1) = ∞

radius of curvature of the right surface (R2) = -12.5 cm (because it is a planoconvex lens with the radius in the direction of the incident rays)

index of refraction (n) = 1.7

(A) we can find the location of the image by applying the formula below

$\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$ where

s' = distance of the image
f = focal length

but we first need to find the focal length before we can apply this formula

$\frac{1}{f} =(n-1)(\frac{1}{R1} -\frac{1}{R2} )$

$\frac{1}{f} =(1.7-1)(\frac{1}{∞} -\frac{1}{-12.5} )$

$\frac{1}{f} =(0.7)(0 + \frac{1}{12.5} )$

$\frac{1}{f} =\frac{0.7}{12.5}$

f = $\frac{12.5}{0.7}$

f = 17.9 cm

now that we have the focal length we can apply $\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$

$\frac{1}{f} - \frac{1}{s}$ =\frac{1}{s'}

$\frac{1}{17.9} - \frac{1}{25}$ =\frac{1}{s'}

$\frac{25 - 17.9}{17.9 x 25} =\frac{1}{s'}$

$\frac{7.1}{447.5} =\frac{1}{s'}$

s' = \frac{447.5}{7.1}[/tex] = 63 cm to the right of the lens

magnification = $\frac{-s'}{s} =\frac{y'}{y}$ where y' is the height of the image, therefore

$\frac{-s'}{s} =\frac{y'}{y}$

$\frac{-63}{25} =\frac{y'}{52.5}$

y' = $\frac{-63}{25}$ x 0.525 = -13.22 cm

therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

(B) if the lens is reversed, the radius of curvatures would be interchanged

radius of curvature of the flat left surface (R1) = ∞

radius of curvature of the right surface (R2) = 12.5 cm

we can find the location of the image by applying the formula below

$\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$ where

s' = distance of the image
f = focal length

but we first need to find the focal length before we can apply this formula

$\frac{1}{f} =(n-1)(\frac{1}{R1} -\frac{1}{R2} )$

$\frac{1}{f} =(1.7-1)(\frac{1}{12.5} -\frac{1}{∞} )$

$\frac{1}{f} =(0.7)( \frac{1}{12.5} - 0)$

$\frac{1}{f} =\frac{0.7}{12.5}$

f = $\frac{12.5}{0.7}$

f = 17.9 cm

now that we have the focal length we can apply $\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$

$\frac{1}{f} - \frac{1}{s}$ =\frac{1}{s'}

$\frac{1}{17.9} - \frac{1}{25}$ =\frac{1}{s'}

$\frac{25 - 17.9}{17.9 x 25} =\frac{1}{s'}$

$\frac{7.1}{447.5} =\frac{1}{s'}$

s' = \frac{447.5}{7.1}[/tex] = 63 cm to the right of the lens

magnification = $\frac{-s'}{s} =\frac{y'}{y}$ where y' is the height of the image, therefore

$\frac{-s'}{s} =\frac{y'}{y}$

$\frac{-63}{25} =\frac{y'}{52.5}$

y' = $\frac{-63}{25}$ x 0.525 = -13.22 cm

therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

7 0

3 years ago