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makvit [3.9K]
3 years ago
9

The second minimum in the diffraction pattern of a 0.11-mm-wide slit occurs at 0.72°. What is the wavelength of the light?

Physics
1 answer:
Anit [1.1K]3 years ago
6 0

Answer:

691.13 nm

Explanation:

d = width of the slit = 0.11 x 10⁻³ m

θ = angle of diffraction pattern = 0.72° degree

λ = wavelength of the light = ?

m = order = 2                              (since second minimum)

for the second minimum diffraction pattern we use the equation

d Sinθ = m λ

Inserting the values

(0.11 x 10⁻³) Sin0.72 = (2) λ

λ = 691.13 x 10⁻⁹ m

λ = 691.13 nm

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Determine the angle of an incline that would yield a constant velocity, given the coefficient of kinetic friction is 0.10.
azamat

Answer:

\theta=5.71^{o}

Explanation:

In order to solve this problem, we mus start by drawing a free body diagram of the given situation (See attached picture).

From the free body diagram we can now do a sum of forces in the x and y direction. Let's start with the y-direction:

\sum F_{y}=0

-W_{y}+N=0

N=W_{y}

so:

N=mgcos(\theta)

now we can go ahead and do a sum of forces in the x-direction:

\sum F_{x}=0

the sum of forces in x is 0 because it's moving at a constant speed.

-f+W_{x}=0

-\mu_{k}N+mg sen(\theta)=0

-\mu_{k}mg cos(\theta)+mg sen(\theta)=0

so now we solve for theta. We can start by factoring mg so we get:

mg(-\mu_{k} cos(\theta)+sen(\theta))=0

we can divide both sides into mg so we get:

-\mu_{k} cos(\theta)+sen(\theta)=0

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\mu_{k} cos(\theta)=sen(\theta)

we now divide both sides of the equation into cos(\theta) so we get:

\mu_{k}=\frac{sen(\theta)}{cos(\theta)}

\mu_{k}=tan(\theta)

so we now take the inverse function of tan to get:

\theta=tan^{-1}(\mu_{k})

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\theta=5.71^{o}

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3 years ago
The equation P^xV^yT^z= constant is Boyle law for what is the values of x,y,z​
Setler79 [48]

Answer:

x = 1, y = 1 and z = 0

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Given equation;

P^x V^y T^z = constant

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PV = constant

From the given equation, the values of x, y and z that will match this law is calculated as follows;

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3 years ago
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