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7nadin3 [17]
3 years ago
13

An experiment based at New Mexico’s Apache Point observatory uses a laser beam to measure the distance to the Moon with millimet

er precision. The laser power is 120 GW, although it’s pulsed on for only 90 ps. The beam emerges from the laser with a diameter of 7.0 mm. It’s then beamed into a telescope aimed at the Moon. When the beam leaves the telescope, it has the telescope’s full 3.5-mm diameter. By the time it reaches the Moon, the beam has expanded to a diameter of 6.5 km.
a. Find the intensity of the beam as it leaves the laser. Express your answer with the appropriate units.
b. Find the intensity of the beam as it leaves the telescope. Express your answer with the appropriate units.
Physics
1 answer:
Anna007 [38]3 years ago
6 0

Answer:

Sorry but i dont know

Explanation:

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A laser beam with a frequency of 180 Hz forms an 8 m standing wave with 10 nodes.
DIA [1.3K]

Answer:33

Explanation:

F = frequency

N =  Node count

w = wave lenght

v = wave velocity

L = distance wave traveled

First find wave length of laser

w = (2/(N))*(L)

w = (2/(10))*(8)

w = 1.6

then using (w), find velocity

V =  (w)(F)

V = (1.6)*(108)

V = 288

Plug in V and the new frequency to solve for new node count

F = NV/2L

(600) = (N)*(288) / 2 * (8)

(N) = 33.33

there are 33 nodes

8 0
3 years ago
Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

4 0
3 years ago
The initial height of the water in a sealed container of diameter 100.0 cm is 5.00 m. The air pressure inside the container is 0
katen-ka-za [31]

Answer:

a)  F = 2.66 10⁴ N, b)   h = 1.55 m

Explanation:

For this fluid exercise we use that the pressure at the tap point is

Exterior

          P₂ = P₀ = 1.01 105 Pa

inside

         P₁ = P₀ + ρ g h

the liquid is water with a density of ρ=1000 km / m³

         P₁ = 0.85   1.01 10⁵ + 1000   9.8  5

         P₁ = 85850 + 49000

         P₁ = 1.3485 10⁵ Pa

the net force is

         ΔP = P₁- P₂

         Δp = 1.3485 10⁵ - 1.01 10⁵

         ΔP = 3.385 10⁴ Pa

Let's use the definition of pressure

         P = Fe / A

         F = P A

the area of ​​a circle is

         A = pi r² = [i d ^ 2/4

let's reduce the units to the SI system

         d = 100 cm (1 m / 100 cm) = 1 m

         F = 3.385 104 pi / 4 (1) ²

         F = 2.66 10⁴ N

b) the height for which the pressures are in equilibrium is

        P₁ = P₂

        0.85 P₀ + ρ g h = P₀

        h = \frac{P_o ( 1-0.850)}{\rho \ g}

        h = \frac{1.01 \ 10^5 ( 1 -0.85)}{1000 \ 9.8}

        h = 1.55 m

4 0
2 years ago
An airplane is cruising at an altitude of several kilometers. The pressure outside the craft is 0.258 atm within the passenger c
garik1379 [7]

Answer:

<em>The speed of the stream of air flowing through the leak is 340.754 m/sec</em>

Explanation:

Carefully applying the Bernoulli's equation the speed of the leak can be obtained. The attached images show step by step explanation of the question, while applying the Bernoulli's equation;

8 0
3 years ago
An empty rubber balloon has a mass of 12.5 g. The balloon is filled with helium at a density of 0.181 kg/m3. At this density the
Taya2010 [7]

Answer:

  • 5.5 N

Explanation:

mass of balloon (m) = 12.5 g = 0.0125 kg

density of helium = 0.181 kg/m^{3}

radius of the baloon (r) = 0.498 m

density of air = 1.29 kg/m^{3}

acceleration due to gravity (g) = 1.29 m/s^{2}

find the tension in the line

the tension in the line is the sum of all forces acting on the line

Tension =buoyant force  + force by helium + force of weight of rubber

force = mass x acceleration

from density = \frac{mass}{volume} ,  mass = density x volume

  • buoyant force =  density x volume x acceleration

        where density is the density of air for the buoyant force

        buoyant force = 1.29 x (\frac{4]{3} x π x 0.498^{3}) x 9.8 = 6.54 N

  • force by helium =  density x volume x acceleration

        force by helium =  0.181 x (\frac{4]{3} x π x 0.498^{3}) x 9.8 = 0.917 N

  • force of its weight = mass of rubber x acceleration

        force of its weight = 0.0125 x 9.8 = 0.1225 N

  • Tension = buoyant force  + force by helium + force of weight of rubber

         the force  of weight of rubber and of helium act downwards, so they      

          carry a negative sign.

  • Tension = 6.54 - 0.917 - 0.1225 = 5.5 N
8 0
3 years ago
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