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Nataly_w [17]
2 years ago
11

Scientists use laser range-finding to measure the distance to the moon with great accuracy. A brief laser pulse is fired at the

moon, then the time interval is measured until the "echo" is seen by a telescope. A laser beam spreads out as it travels because it diffracts through a circular exit as it leaves the laser. In order for the reflected light to be bright enough to detect, the laser spot on the moon must be no more than 1.0 km in diameter. Staying within this diameter is accomplished by using a special large-diameter laser. If λ=532 nm, what is the minimum diameter of the circular opening from which the laser beam emerges? The earth-moon distance is 384,000 km.
Physics
1 answer:
Fofino [41]2 years ago
5 0

Answer:

d = 2,042 10-3 m

Explanation:

The laser diffracts in the circular slit, so the process equation is

      d sin θ= m λ

The first diffraction minimum occurs for m = 1

We can use trigonometry in the mirror

        tan θ = Y / L

Where L is the distance from the Moon to Earth

Since the angle is extremely small

           tan θ = sin θ / cos θ

           Cos θ = 1

           tant θ = sin θ = y / L

We replace

           d y / L = λ

           d = λ L / y

Let's calculate

           d = 532 10⁻⁹ 3.84 10⁶/1 10³

           d = 2,042 10-3 m

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A sealed tank containing seawater to a height of 10.5 mm also contains air above the water at a gauge pressure of 2.95 atmatm. W
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The water is flowing at the rate of 28.04 m/s.

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To determine the speed of the water, apply Bernoulli's equation;

P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2

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Substitute in these values and the Bernoulli's equation will reduce to;

P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 =  P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 + \frac{1}{2}\rho (0)^2 =  P_2 + \rho g(0) + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 =  P_2 + \frac{1}{2}\rho v_2^2\\\\P_{gauge} + P_{atm} + \rho gz_1 = P_{atm} + \frac{1}{2}\rho v_2^2\\\\P_{gauge} +  \rho gz_1 =  \frac{1}{2}\rho v_2^2\\\\v_2^2 = \frac{2(P_{gauge} +  \rho gz_1)}{\rho} \\\\v_2 = \sqrt{ \frac{2(P_{gauge} +  \rho gz_1)}{\rho} }

where;

\rho is the density of seawater = 1030 kg/m³

v_2 = \sqrt{ \frac{2(2.95*101325 \ + \  1030*9.8*10.5 )}{1030} }\\\\v_2 = 28.04 \ m/s

Therefore, the water is flowing at the rate of 28.04 m/s.

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3 years ago
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