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1 year ago

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What is the angle of the 2nd order bright fringe produced by two slits that are 8.25x10-5m apart if the wavelength of the incide

nt light is 4.50x10-7m?0.0109⁰0.625°91.7°1.60⁰

1 answer:

dexar [7]1 year ago

5 0

In order to calculate the angle, we can use the formula below for a constructive interference (the interference is constructive because the fringe is bright):

$d\sin\theta=m\lambda$

Where d is the distance between the slits, m is the order of the interference and lambda is the wavelength.

So, using d = 8.25 * 10^-5, m = 2 and lambda = 4.5 * 10^-7, we have:

$\begin{gathered} 8.25\cdot10^{-5}\cdot\sin\theta=2\cdot4.5\cdot10^{-7}\\ \\ \sin\theta=\frac{9\cdot10^{-7}}{8.25\cdot10^{-5}}\\ \\ \sin\theta=1.091\cdot10^{-2}\\ \\ \theta=0.625° \end{gathered}$

Therefore the correct option is the second one.

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9. A 5.0 kg block on an inclined plane is acted upon by a horizontal force of 100 N shown in the figure below. The coefficient o

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Answer:

A: The acceleration is 7.7 m/s up the inclined plane.

B: It will take the block 0.36 seconds to move 0.5 meters up along the inclined plane

Explanation:

Let us work with variables and set

$m=5kg\\\\F_H=100N\\\\\mu=0.3\\\\\theta=37^o.$

As shown in the attached free body diagram, we choose our coordinates such that the x-axis is parallel to the inclined plane and the y-axis is perpendicular. We do this because it greatly simplifies our calculations.

Part A:

From the free body diagram we see that the total force along the x-axis is:

$F_{tot}=mg*sin(\theta)+F_s-F_Hcos(\theta).$

Now the force of friction is $F_s=\mu*N,$ where $N$ is the normal force and from the diagram it is $F_y=mg*cos(\theta).$

Thus $F_s=\mu*N=\mu*mg*cos(\theta).$

Therefore,

$F_{tot}=mg*sin(\theta)+\mu*mg*cos(\theta)-F_Hcos(\theta)\\\\=mg(sin(\theta)+\mu*cos(\theta))-F_Hcos(\theta).$

Substituting the value for $F_H,m,\mu, and \:\theta$ we get:

$F_{tot}= -38.63N.$

Now acceleration is simply

$a=\frac{F_H}{m} =\frac{-38.63N}{5kg} =-7.7m/s.$

The negative sign indicates that the acceleration is directed up the incline.

Part B:

$d=\frac{1}{2} at^2$

Which can be rearranged to solve for t:

$t=\sqrt{\frac{2d}{a} }$

Substitute the value of $d=0.50m$ and $a=7.7m/s$ and we get:

$t=0.36s.$

which is our answer.

Notice that in using the formula to calculate time we used the positive value of $a$ , because for this formula absolute value is needed.

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3 years ago

A circular cross section, d = 25 mm, experiences a torque load, T = 25 N·m, and a shear force, V = 85 kN. Calculate the shear st

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$\tau$ = shear stress at a distance 'r' from the center

T = is the applied torque

$I_{p}$ = polar moment of inertia of the section

r = radial distance from the center

Thus we can see that if a point is located at center i.e r = 0 there will be no shearing stresses at the center due to torque.

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2 years ago

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Answer:

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