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MakcuM [25]
3 years ago
9

After a laser beam passes through two thin parallel slits, the first completely dark fringes occur at 19.0 with the original dir

ection of the beam, as viewed on a screen far from the slits. (a) What is the ratio of the distance between the slits to the wavelength of the light illuminating the slits
Physics
1 answer:
Katarina [22]3 years ago
8 0

Answer:

$\frac{d}{\lambda} = 1.54$

Explanation:

Given :

The first dark fringe is for m = 0

$\theta_1 = \pm 19^\circ$

Now we know for a double slit experiments , the position of the dark fringes is give by :

$d \sin \theta=\left(m+\frac{1}{2}\right) \lambda$

The ratio of distance between the two slits, d to the light's wavelength that illuminates the slits, λ :

$d \sin \theta=\left(\frac{1}{2}\right) \lambda$     (since, m = 0)

$d \sin \theta=\frac{\lambda}{2}$

$\frac{d}{\lambda} = \frac{1}{2 \sin \theta}$

$\frac{d}{\lambda} = \frac{1}{2 \sin 19^\circ}$

$\frac{d}{\lambda} = 1.54$

Therefore, the ratio is $\frac{1}{1.54}$  or 1 : 1.54

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A 26.2-kg dog is running northward at 3.02 m/s, while a 5.30-kg cat is running eastward at 2.74 m/s. Their 65.1-kg owner has the
REY [17]

Answer:

Angle with the +x axis is θ = 79.599degree

Then the velocity of owner = 1.235m/s

Explanation:

Given that the mass of dog is m1 =26.2 kg

velocity of dog is u1 = 3.02 m/s (north)

mass of cat is m2 = 5.3 kg

velocity is u2 = 2.74 m/s (east )

Mass of owner is M = 65.1 kg

Consider the east direction along +x axis andnorth along +y

momentum of dog is Py = m1 x u1

= 79.124 kg.m/s (j)

momentum of cat is Px = m2 x u2

= 14.522 kg.m/s (i)

Then the net magnitude of momentum is P = (Px2 + Py2)1/2

= 80.445

Angle with the +x axis is θ =tan-1(Py / Px ) = 79.599 degree

Then the velocity of owner is v = P / M = 1.235 m/s

3 0
3 years ago
A scuba diver at 70 m below the surface of a lake, where the temperature is 4 degrees C, releases an air bubble with a volume of
posledela

Answer:

121.3 cm^3

Explanation:

P1 = Po + 70 m water pressure (at a depth)

P2 = Po (at the surface)

T1 = 4°C = 273 + 4 = 277 K

V1 = 14 cm^3

T2 = 23 °C = 273 + 23 = 300 K

Let the volume of bubble at the surface of the lake is V2.

Density of water, d = 1000 kg/m^3

Po = atmospheric pressure = 10^5 N/m^2

P1 = 10^5 + 70 x 1000 x 10 = 8 x 10^5 N/m^2

Use the ideal gas equation

\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

By substituting the values, we get

\frac{8\times 10^5\times 14}{277}=\frac{10^{5} \timesV_{2}}{300}

V2 = 121.3 cm^3

Thus, the volume of bubble at the surface of lake is 121.3 cm^3.

6 0
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6 0
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What kind of research in teaching of physical science can be done???​
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Answer:

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Explanation:

3 0
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