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Sholpan [36]
3 years ago
9

Irina finds an unlabeled box of fine needles, and wants to determine how thick they are. A standard ruler will not do the job, a

s each needle is less than a millimeter thick. So, to find the thickness, she uses a needle to poke a hole in a piece of brown construction paper. Then, she positions a 640 nm laser pointer to shine through the hole and project a circular diffraction pattern on a wall 21.7 m away. She then uses her ruler to measure that the central bright circle is 14.2 cm in diameter. What diameter does Irina calculate for the needle?
Physics
1 answer:
Vsevolod [243]3 years ago
8 0

Answer:

d=\frac{1.22\times 640\times 10^{-9}\times 21.7}{7.35\times 10^{-2}}=2305.21\times 10^{-7}m

Explanation:

The expression which represent the first diffraction minima by a circular aperture is given by d sin\Theta =1.22\lambda--------eqn 1

The angle through which the first minima is diffracted is given by tan\Theta =\frac{y_1}{D}---------eqn 2

As \Theta is very small so we can write sin\Theta =tan\Theta

So from eqn 1 and eqn 2 we can write

y_1=\frac{1.22\lambda D}{d}--------eqn 3

Here y_1 is the position of first maxima D is the distance of screen from the circular aperture d is the diameter of aperture

It is given that diameter of circular aperture is 14.7 cm so y_1=\frac{14.7}{2}=7.35 \ cm

Now putting all these value in eqn 3

d=\frac{1.22\lambda D}{y_1}

d=\frac{1.22\times 640\times 10^{-9}\times 21.7}{7.35\times 10^{-2}}=2305.21\times 10^{-7}m

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A disk with mass 1.64 kg and radius 0.61 meters is spinning counter-clockwise with an angular velocity of 17.6 rad/s. A rod of m
Masja [62]

Answer:

The loss in rotational kinetic energy due to the collision is 36.585 J.

Explanation:

Given;

mass of the disk, m₁ = 1.64 kg

radius of the disk, r = 0. 61 m

angular velocity of the disk, ω₁ = 17.6 rad/s

mass of the rod, m₂ = 1.51 kg

length of the rod, L = 1.79 m

angular velocity of the rod, ω₂ =  5.12 rad/s (clock-wise)

let the counter-clockwise be the positive direction

let the clock-wise be the negative direction

The common final velocity of the two systems after the collision is calculated by applying principle of conservation of angular momentum ;

m₁ω₁  + m₂ω₂ = ωf(m₁ + m₂)

where;

ωf is the common final angular velocity

1.64 x 17.6    + 1.51(-5.12) = ωf(1.64 + 1.51)

21.1328 = ωf(3.15)

ωf = 21.1328 / 3.15

ωf = 6.709 rad/s

The moment of inertia of the disk is calculated as follows;

I_{disk} = \frac{1}{2} mr^2\\\\I_{disk}  = \frac{1}{2} (1.64)(0.61)^2\\\\I_{disk}  = 0.305 \ kgm^2

The moment of inertia of the rod about its center is calculated as follows;

I_{rod} = \frac{1}{12} mL^2\\\\I_{rod} = \frac{1}{12} \times 1.51 \times 1.79^2\\\\I _{rod }= 0.4032\ kgm^2

The initial rotational kinetic energy of the disk and rod;

K.E_i = \frac{1}{2} I_{disk}\omega _1 ^2 \ \ + \ \  \frac{1}{2} I_{rod}\omega _2 ^2 \\\\K.E_i=  \frac{1}{2} (0.305)(17.6) ^2 \ \ + \ \  \frac{1}{2} (0.4032)(-5.12) ^2\\\\K.E_i = 52.523 \ J

The final rotational kinetic energy of the disk-rod system is calculated as follows;

K.E_f = \frac{1}{2} I_{disk}\omega _f ^2 \ \ + \ \  \frac{1}{2} I_{rod}\omega _f ^2\\\\K.E_f = \frac{1}{2} \omega _f ^2(I_{disk} + I_{rod})\\\\K.E_f = \frac{1}{2} (6.709) ^2(0.305+ 0.4032)\\\\K.E_f = 15.938 \ J

The loss in rotational kinetic energy due to the collision is calculated as follows;

\Delta K.E = K.E_f \ - \ K.E_i\\\\\Delta K.E = 15.938 J  \ - \ 52.523 J\\\\\Delta K.E = - 36.585 \ J

Therefore, the loss in rotational kinetic energy due to the collision is 36.585 J.

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2 years ago
The organ system that moves blood around your body?
dusya [7]
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3 years ago
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Answer:

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Referring to the diagram below which represents the rock cycle, we will find that applying heat and pressure on sedimentary rocks would convert these rocks into metamorphic rocks.

Based on the above, the best choice would be:
<span>d. Heat and pressure can change sedimentary rocks into metamorphic rocks.</span>

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