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Alik
3 months 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:
Ostrovityanka [3.2K]3 months ago
6 0

Answer:

Unfortunately, I do not possess the information

Explanation:

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A 5.00-A current runs through a 12-gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5 * 1028 free ele
Maru [3345]

Answer:

a)n= 3.125 x 10^{19 electrones.

b)J= 1.515 x 10^{6 A/m²

c)V_{d =1.114 x 10^{4m/s

d) ver explicación

Explanation:

La corriente 'I' = 5A =>5C/s

diámetro 'd'= 2.05 x 10^{-3 m

radio 'r' = d/2 => 1.025 x 10^{-3 m

número de electrones 'n'= 8.5 x 10^{28}

a) La cantidad de electrones que pasan por la bombilla cada segundo se determina mediante:

I= Q/t

Q= I x t => 5 x 1

Q= 5C

Como sabemos que: Q= ne

donde e es la carga del electrón, es decir, 1.6 x 10^{-19C

n= Q/e => 5/ 1.6 x 10^{-19

n= 3.125 x 10^{19 electrones.

b) La densidad de corriente 'J' en el cable se calcula como

J= I/A => I/πr²

J= 5 / (3.14 x (1.025x 10^{-3)²)

J= 1.515 x 10^{6 A/m²

c) La velocidad típica 'V_{d' de un electrón se expresa como:

V_{d = \frac{J}{n|q|}

    =1.515 x 10^{6 / 8.5 x 10^{28} x |-1.6 x 10^{-19|

V_{d =1.114 x 10^{4m/s

d) De acuerdo con estas ecuaciones,

J= I/A

V_{d = \frac{J}{n|q|} =\frac{I}{nA|q|}

Si utilizaras un cable de doble diámetro, ¿cuáles de las respuestas anteriores cambiarían? ¿Aumentarían o disminuirían?

5 0
4 months ago
Read 2 more answers
the minute hand on a clock is 9 cm long and travels through an arc of 252 degrees every 42 minutes. To the nearest tenth of a ce
Ostrovityanka [3204]

Answer:

The distance covered by the minutes hand is 39.60 cm.

Explanation:

Note: A clock has a circular shape, where the minutes hand acts as the radius, and its motion creates an arc.

Length of an arc is calculated as ∅/360(2πr)

L = ∅/360(2πr).................... Equation 1π

Here, L represents the arc’s length, ∅ is the angle made by the arc, and r is the arc’s radius.

Given: ∅ = 252°, r = 9 cm, π = 3.143.

Upon substituting these values into equation 1,

L = 252/360(2×3.143×9)

L = 0.7×2×3.143×9

L = 39.60 cm.

Thus, the distance traversed by the minutes hand is 39.60 cm.

4 0
4 months ago
Which statements describe properties of stars check all that apply
Keith_Richards [3271]

Answer:

Stars generate energy by the process of nuclear fusion.

They are large entities composed of gaseous elements.

The main constituents of stars are hydrogen and helium.

Explanation:

Stars are colossal objects with extensive gravitational forces causing them to contract, which allows fusion to take place: the atomic nuclei in the star's core are drawn very close together due to gravity and elevated temperatures, leading to the fusion reaction. This fusion serves as the energy output for a star.

Conversely, it is true that stars predominantly consist of hydrogen and helium (two hydrogen nuclei can fuse to become helium), which implies that a star is essentially an enormous ball of gas without a solid surface suitable for standing on.

As for the presence of water on a star, it is simply impossible. The extreme temperatures found in stars are far too high for water to exist in any liquid state on their surfaces.

7 0
3 months ago
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