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lutik1710 [3]
3 years ago
5

(Solidification) You are performing a double slit experiment very similar to the one from DLby shining a laser on two narrow sli

ts spaced 7.5 × 10-5meters apart. However, by placing a piece of crystal in one of the slits, you are able to make it so that the rays of light that travel through the two slits are πout of phasewith each other (that is to say,Δφ!=휋).If you observe that on a screen placed 4meters from the two slits that the distance between the bright spot closest to the center of the interference pattern and the center of the pattern is 1.5 cm, what is the wavelength of the laser? (Just as in DL, you may use the small angle approximation sinθ ≈ θ ≈ tanθ.)
Physics
1 answer:
anzhelika [568]3 years ago
3 0

Answer:

The wavelength of the laser, λ = 5.625 * 10⁻⁷ m

Explanation:

Separation of the narrow slits, d = 7.5 * 10⁻⁵ m

The distance between the screen and the two slits, d = 4m

The distance between the bright spot and the center of the pattern, Y = 1.5 cm

Y = 1.5 * 10⁻² m

To calculate the wavelength, λ, of the laser we will use the relationship:

Y = \frac{\lambda L}{2d} \\1.5 * 10^{-2}  = \frac{\lambda * 4}{2*7.5*10^{-5} }\\\lambda = \frac{1.5 * 10^{-2} * 15 * 10^{-5} }{4}

λ = 5.625 * 10⁻⁷ m

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Two balls of clay, with masses M1 = 0.49 kg and M2 = 0.47 kg, are thrown at each other and stick when they collide. Mass 1 has a
malfutka [58]

Answer:

a) p_i=1.568\hat{i}+0.752 \hat{j}

b) v_{fx}=1.668\ m.s^{-1}

c) v_{fy}=0.7999\ m.s^{-1}

Explanation:

Given masses:

m_1=0.49\ kg

m_2=0.47\ kg

Velocity of mass 1, v_1=3.2 \hat{i}\ m.s^{-1}

Velocity of mass 2, v_2=1.6 \hat{j}\ m.s^{-1}

a)

Initial momentum:

p_i=m_1.v_1+m_2.v_2

p_i=0.49\times 3.2 \hat{i}+0.47\times 1.6 \hat{j}

p_i=1.568\hat{i}+0.752 \hat{j}

b)

magnitude of initial momentum:

p_i=\sqrt{1.568^2+0.752 ^2}

p_i=1.739\ kg.m.s^{-1}

From the conservation of momentum:

p_f=p_i

m_f.v_f=1.739

v_f=\frac{1.739}{0.49+0.47}

v_f=1.85\ m.s^{-1} is the magnitude of final velocity.

Direction of final velocity will be in the direction of momentum:

tan\theta=\frac{0.752 }{1.568}

\theta=25.62^{\circ}

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v_{fx}=1.668\ m.s^{-1}

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v_{fy}=1.85\ sin 25.62^{\circ}

v_{fy}=0.7999\ m.s^{-1}

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3 years ago
A ball is dropped from rest from the top of a building of height h. At the same instant, a second ball is projected vertically u
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Answer:

a) t = \sqrt{\frac{h}{2g}}

b) Ball 1 has a greater speed than ball 2 when they are passing.

c) The height is the same for both balls = 3h/4.

Explanation:

a) We can find the time when the two balls meet by equating the distances as follows:

y = y_{0_{1}} + v_{0_{1}}t - \frac{1}{2}gt^{2}  

Where:

y_{0_{1}}: is the initial height = h

v_{0_{1}}: is the initial speed of ball 1 = 0 (it is dropped from rest)

y = h - \frac{1}{2}gt^{2}     (1)

Now, for ball 2 we have:

y = y_{0_{2}} + v_{0_{2}}t - \frac{1}{2}gt^{2}    

Where:

y_{0_{2}}: is the initial height of ball 2 = 0

y = v_{0_{2}}t - \frac{1}{2}gt^{2}    (2)

By equating equation (1) and (2) we have:

h - \frac{1}{2}gt^{2} = v_{0_{2}}t - \frac{1}{2}gt^{2}

t=\frac{h}{v_{0_{2}}}

Where the initial velocity of the ball 2 is:

v_{f_{2}}^{2} = v_{0_{2}}^{2} - 2gh

Since v_{f_{2}}^{2} = 0 at the maximum height (h):

v_{0_{2}} = \sqrt{2gh}

Hence, the time when they pass each other is:

t = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}

b) When they are passing the speed of each one is:

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v_{f_{1}} = - gt = -g*\sqrt{\frac{h}{2g}} = - 0.71\sqrt{gh}

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v_{f_{2}} = v_{0_{2}} - gt = \sqrt{2gh} - g*\sqrt{\frac{h}{2g}} = (\sqrt{1} - \frac{1}{\sqrt{2}})*\sqrt{gh} = 0.41\sqrt{gh}

Therefore, taking the magnitude of ball 1 we can see that it has a greater speed than ball 2 when they are passing.

c) The height of the ball is:

For ball 1:

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For ball 2:

y_{2} = v_{0_{2}}t - \frac{1}{2}gt^{2} = \sqrt{2gh}*\sqrt{\frac{h}{2g}} - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h

Then, when they are passing the height is the same for both = 3h/4.

I hope it helps you!                  

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