Ask question

Ask question

All categories

2 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.

1 answer:

Anna007 [38]2 years ago

6 0

Answer:

Sorry but i dont know

Explanation:

You might be interested in

When a potential difference of 12 V is applied to a wire 6.7 m long and 0.31 mm in diameter the result is an electric current of

anastassius [24]

The wire needs to be sauderwired to be connected back into place to get energy into column so it came function properly again!

7 0

2 years ago

Solve the inequality 2(n+3) – 4<6. Then graph<br> the solution.

Aloiza [94]

The solution is 22 2(n+3)-4&6

6 0

2 years ago

A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0 m/s2. Secret agent Austin Powers jumps on ju

denpristay [2]

Answer:

a) $h=250\ m$

b) $\Delta h=0.0835\ m$

Explanation:

Given:

upward acceleration of the helicopter, $a=5\ m.s^{-2}$

time after the takeoff after which the engine is shut off, $t_a=10\ s$

a)

<u>Maximum height reached by the helicopter:</u>

using the equation of motion,

$h=u.t+\frac{1}{2} a.t^2$

where:

u = initial velocity of the helicopter = 0 (took-off from ground)

t = time of observation

$h=0+0.5\times 5\times 10^2$

$h=250\ m$

b)

time after which Austin Powers deploys parachute(time of free fall), $t_f=7\ s$

acceleration after deploying the parachute, $a_p=2\ m.s^{-2}$

<u>height fallen freely by Austin:</u>

$h_f=u.t_f+\frac{1}{2} g.t_f^2$

where:

$u=$ initial velocity of fall at the top = 0 (begins from the max height where the system is momentarily at rest)

$t_f=$ time of free fall

$h_f=0+0.5\times 9.8\times 7^2$

$h_f=240.1\ m$

<u>Velocity just before opening the parachute:</u>

$v_f=u+g.t_f$

$v_f=0+9.8\times 7$

$v_f=68.6\ m.s^{-1}$

<u>Time taken by the helicopter to fall:</u>

$h=u.t_h+\frac{1}{2} g.t_h^2$

where:

$u=$ initial velocity of the helicopter just before it begins falling freely = 0

$t_h=$ time taken by the helicopter to fall on ground

$h=$ height from where it falls = 250 m

now,

$250=0+0.5\times 9.8\times t_h^2$

$t_h=7.1429\ s$

From the above time 7 seconds are taken for free fall and the remaining time to fall with parachute.

<u>remaining time,</u>

$t'=t_h-t_f$

$t'=7.1428-7$

$t'=0.1428\ s$

<u>Now the height fallen in the remaining time using parachute:</u>

$h'=v_f.t'+\frac{1}{2} a_p.t'^2$

$h'=68.6\times 0.1428+0.5\times 2\times 0.1428^2$

$h'=9.8165\ m$

<u>Now the height of Austin above the ground when the helicopter crashed on the ground:</u>

$\Delta h=h-(h_f+h')$

$\Delta h=250-(240.1+9.8165)$

$\Delta h=0.0835\ m$

5 0

2 years ago

A. An automobile mass is 3.5 x103 kg. If the forward thrust (Fnet)

katrin [286]

Answer:

a = 0.8 m/s^2

Explanation:

Force equation: F = ma

F = ma -> a = F/m = 2.8*10^3 N / 3.5*10^3 kg = 0.8 m/s^2

8 0

2 years ago

Two poles are connected by a wire that is also connected to the ground. The first pole is 20ft tall and the second pole is 10ft

Natasha2012 [34]

Answer: 31.6ft

Explanation:

Check the attachment for the diagram.

According to the right angle triangle AEC, we will use Pythagoras theorem to get |AC|. Note that |AE| = |AB| - |CD|

that is 20ft - 10ft = 10ft

According to the theorem, the square of the sum of the adjacent side and the opposite side is equal to the square of the hypotenuse.

|AE|^2 + |EC|^2 = |AC|^2

10^2 + 30^2 = |AC|^2

100 + 900 = |AC|^2

|AC| = √1000

|AC| = 31.6ft

Therefore, the wire should be anchored 31.6ft to the ground to minimize the amount of wire needed.

6 0

3 years ago