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stealth61 [152]

3 years ago

8

A series of bright fringes appears on the viewing screen of a Young's double-slit experiment. Suppose you move from one bright f

ringe to the next one farther out from the central bright fringe. Does the difference in path lengths of the light waves increase or decrease, and by what amount does it change? Express your answer in terms of the wavelength λ of the light.

1 answer:

goblinko [34]3 years ago

6 0

Ahahaha fidjdsd skssjsjsbs SJSU’s

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A transverse wave on a string is described by the following wave function.y = (0.090 m) sin (px/11 + 4pt)(a) Determine the trans

alukav5142 [94]

Explanation:

(a) It is known that equation for transverse wave is given as follows.

y = $(0.09 m)sin(\pi \frac{x}{11} + 4 \pi t)$

Now, we will compare above equation with the standard form of transeverse wave equation,

y = $A sin(kx + \omega t)$

where, A is the amplitude = 0.09 m

k is the wave vector = $\frac{\pi}{11}$

$\omega$ is the angular frequency = $4\pi$

x is displacement = 1.40 m

t is the time = 0.16 s

Now, we will differentiate the equation with respect to t as follows.

The speed of the wave will be:

v(t) = $\frac{dy}{dt}$

v(t) = $A \omega cos(kx + \omega t)$

v(t) = $(0.09 m)(4\pi) cos(\frac{\pi \times 1.4}{11} + 4 \pi \times 0.16)$

v(t) = -0.84 m/s

The acceleration of the particle in the location is

a(t) = $\frac{dv}{dt}$

a(t) = $-A \omega 2sin(kx + \omega t)$

a(t) = $-(0.09 m)(4 \pi)2 sin(\frac{\pi \times 1.4}{11} + 4\pi \times 0.16)$

a(t) = -9.49 $m/s^{2}$

Hence, the value of transverse wave is 0.84 m/s and the value of acceleration is 9.49 $m/s^{2}$ .

(b) Wavelength of the wave is given as follows.

$\lambda = \frac{2\pi}{k}$

$\lambda = (frac{2\pi}{\frac{\pi}{11})
$

$\lambda$ = 22 m

The period of the wave is

T = $\frac{2 \pi}{\omega}$

T = $\frac{2 \pi}{4 \pi}$

= 0.5 sec

Now, we will calculate the speed of propagation of wave as follows.

v = $\frac{\lambda}{T}$

= $\frac{22 m}{0.5 s}$

= 44 m/s

therefore, we can conclude that wavelength is 22 m, period is 0.5 sec, and speed of propagation of wave is 44 m/s.

7 0

3 years ago

The average speed of a nitrogen molecule in air is about 6.70×102 m/s, and its mass is 4.68×10-26 kg.

Otrada [13]

Answer:

a) a = 3.06 10¹⁵ m / s , b) F= 1.43 10⁻¹⁰ N, c) F_total = 14.32 10⁻²⁶ N

Explanation:

This exercise will average solve using the moment relationship.

a ) let's use the relationship between momentum and momentum

I = ∫ F dt = Δp

F t = m $v_{f}$ - m v₀

F = m (v_{f} -v₀o) / t

in the exercise indicates that the speed module is the same, but in the opposite direction

F = m (-2v) / t

if we use Newton's second law

F = m a

we substitute

- 2 mv / t = m a

a = - 2 v / t

let's calculate

a = - 2 4.59 10²/3 10⁻¹³

a = 3.06 10¹⁵ m / s

b) F= m a

F= 4.68 10⁻²⁶ 3.06 10¹⁵

F= 1.43 10⁻¹⁰ N

c) if we hit the wall for 1015 each exerts a force F

F_total = n F

F_total = n m a

F_total = 10¹⁵ 4.68 10⁻²⁶ 3.06 10¹⁵

F_total = 14.32 10⁻²⁶ N

8 0

3 years ago

Analogy: atom is to element as ____ is to compound

Masja [62]

Molecule is the answer to what you are looking for

5 0

3 years ago

We can see Objects because of

Amiraneli [1.4K]

Answer:

I believe it's A.)

Explanation:

Although light comes into our atmosphere through refraction, it reaches our eyes only through reflection from objects. So when light rays reflect off an object and enter the eyes through the cornea you can then see that object.

Hope this helps you out : )

7 0

3 years ago

In which of the following would the particles move most rapidly? a. ice at -20 °C b. water at 20 °C c. steam at 110 °C d. boilin

Artemon [7]

Answer:

C. steam at 110

Explanation:

3 0

3 years ago