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Contact [7]
4 years ago
9

Two sinusoidal waves with identical wavelengths and amplitudes travel in opposite directions along a string with a speed of 10 c

m/s. If the time interval between instants when the string is flat is 0.50 s, what is the wavelength of the waves?
Physics
1 answer:
andrey2020 [161]4 years ago
5 0

Answer:

The wave length of the string = 10 cm or 0.1 m

Explanation:

Wave: A wave can be defined as a disturbance, which travels through a medium and transfer energy from one point to another, without causing any permanent displacement of the medium itself.

 The velocity, frequency and wavelength of a wave is related as shown,

v = λf ........................ Equation 1

Where v = velocity of the wave, λ = wave length, f = frequency.

λ = v/f ..................... Equation 2.

Given: v = 10 cm/s,

T = 2t, where t = 0.5 s, (This is half of one cycle)  

T = 2×0.5 = 1 s,

But T = 1/f = 1/1 = 1 Hz.

Substituting these values into equation 2

λ  = 10(1)

λ  = 10 cm. or 0.1 m

Thus the wave length of the waves = 10 cm or 0.1 m

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A laser beam of unknown wavelength passes through adiffraction grating having 5510 lines/cm after striking itperpendicularly. Ta
Ivahew [28]

Answer:

Explanation:

distance between two slit  d = \frac{1 \times 10^{-2}}{5510}

d = 18.15 x 10⁻⁷ m

Let wave length of light λ

formula for  position of  first pair of bright spot

Tanθ = λ / d , λ is wave length of light and d is distance between two slit .

Tan 15.4 = \frac{\lambda}{18.15\times10^{-7}}

λ = Tan 15.4 x 18.15 x 10⁻⁷

=5 x 10⁻⁷ m

If θ be the position of next bright spot

Tanθ = 2 λ / d

= \frac{2 \times\ 5\times\ 10^ {-7}}{18.15\times10^{-7}}

=\frac{2\times5}{18.5}

θ = 28.4 degree .

7 0
3 years ago
The electric field strength is 5.50×10^4 N/C inside a parallel-plate capacitor with a 2.50 mm spacing. A proton is released from
valentina_108 [34]

Answer:

v=1.6\times10^{5}m/s

Explanation:

The equation that relates electric field strength and force that a charge experiments by it is F=qE.

Newton's 2nd Law states F=ma, which for our case will mean:

a=\frac{F}{m}=\frac{qE}{m}

We know from accelerated motion that v^2=v_0^2+2ad

In our case the proton is released from rest, so v_0=0m/s and we get v=\sqrt{2ad}

Substituting, we get our final velocity equation:

v=\sqrt{\frac{2qEd}{m}}

For the <em>proton </em>we know that q=1.6\times10^{-19}C and m=1.67\times10^{-27}kg. Writing in S.I. d=0.0025m, we obtain:

v=\sqrt{\frac{2(1.6\times10^{-19}C)(5.5\times10^{4}N/C)(0.0025m)}{(1.67\times10^{-27}kg)}}=162318.5m/s=1.6\times10^{5}m/s

6 0
3 years ago
Which shows zero as significant digits? Check all that apply.
melisa1 [442]

Answer:

Explanation:

0.02

0.604

70,230

becuase if you take away the 0 then it would change the value of the number

8 0
3 years ago
A small block has constant acceleration as it slides down a frictionless incline. The block is released from rest at the top of
Natasha2012 [34]

Answer:

   h ’= 0.51356 m

Explanation:

For this exercise we can use conservation of energy

starting point. Highest point of the trajectory

         Em₀ = m g h

         sin θ = h / L

         h = L sin θ

         Em₀ = m g L sin  θ

final point. Lowest point of the trajectory

         Em_f = K = ½ mv²

as there is no friction, energy is conserved

        Em₀ = Emf

        m g L sin  θ  = ½ m v²

        sin  θ  = ½ v² / gL

        sin  θ  = ½ 3.80² / (9.8  6.60)

        sin  θ  = 0.111626

         tea = 6.41

They ask us for the speed for L ’= 4.60 m

let's find the height

         sin 6.41 = h '/ L'

           

h ’= L’ sin 6.41

         h ’= 4.60 sin 6.41

         h ’= 0.51356 m

we use conservation of energy

          m g h ’= ½ m v’2

          v ’= √ (2gh’)

          v ’= √ (2 9.8 0.51356)

          v ’= 3.173 m /s²

6 0
3 years ago
In projectile motion, why don't we add acceleration due to gravity on the formula of horizontal motion​
Basile [38]

Answer:

Horizontal motion is free from gravity which is zero

8 0
2 years ago
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