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14 days ago

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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:

Ostrovityanka [2.2K]14 days ago

6 0

Answer:

Unfortunately, I do not possess the information

Explanation:

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Simone is walking her dog on a leash. The dog is pulling with a force of 32 N to the right and Simone is pulling backward with a

ValentinkaMS [2425]

Conclusion:

The total net force acting on the objects is 16 N, directed towards the right.

Clarification:

It is stated that,

The force exerted by the dog, $F_1 = 32\ N$ (to the right)

The force exerted by Simone, $F_2 = -16\ N$ (backward)

Here, assume the backward direction is negative and the right direction is positive.

The net force will move in the direction where the larger force is present. The net force can be calculated as:

$F=F_1+F_2$

$F=32+(-16)$

F = 16 N

Thus, the net force amounts to 16 N, acting towards the right.

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24 days ago

Read 2 more answers

A spin bike has a flywheel in two parts—a 12.5 kg disk with radius 0.23 m, and a 7.0 kg ring with mass concentrated at the outer

Keith_Richards [2263]

Response:

Clarification:

Provided

weight of disk $m=12.5 kg$

diameter of disc $R=0.23 m$

weight of ring $m_r=7 kg$

Force $F=9.7 N$

$N=180 rpm$

$\omega =\frac{2\pi N}{60}$

$\omega =6\pi rad/s$

Overall moment of inertia

=Disc's moment of inertia +Ring's Moment of Inertia

$=0.5\cdot 12.5\times 0.23^2+7\times 0.23^2$

$=13.25\times 0.23^2=0.7009 kg-m^2$

At this point, Torque is $T=F\times R=I\cdot \alpha$

$9.7\times 0.23=0.7\times \alpha$

$\alpha =3.18 rad/s^2$

Utilizing $\omega _f=\omega +\alpha t$

$\omega _f=0$ in this scenario

$0=6\pi -3.18\times t$

$t=\frac{6\pi }{3.18}$

$t=5.92 s$

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1 day ago

Ability of the muscles to function effectively and efficiently without undue fatigue

Keith_Richards [2263]

Response:

Physical well-being

Clarification:

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1 month ago

Justine is ice-skating at the Lloyd Center what is her final velocity if she accelerates at a rate of 2.0 meters per second for

Ostrovityanka [2208]

2*3.5 = 7m/s

You need to multiply the acceleration by the time (which must both be in seconds; if not, convert them to the same units).

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9 days ago

A small block of mass 200 g starts at rest at A, slides to B where its speed is vB=8.0m/s,vB=8.0m/s, then slides along the horiz

serg [2593]

Answer

Data provided:

mass of the block = 200 g = 0.2 Kg

Velocity at A = 0 m/s

Velocity at B = 8 m/s

distance of slide = 10 m

height of the block = 4 m

calculation for the block's potential energy

P = m g h

P = 0.2 x 9.8 x 4

P = 7.84 J

kinetic energy calculated as

$KE = \dfrac{1}{2}mv^2$

$KE = \dfrac{1}{2}\times 0.2 \times 7.84^2$

$KE =6.14 J$

Work done = P - KE

work = 7.84 - 6.14

work = 1.7 J

b) using the formula v² = u² + 2 a s

0 = 8² - 2 x a x 10

a = 3.2 m/s²

ma - μ mg = 0

$\mu = \dfrac{a}{g}$

$\mu = \dfrac{3.2}{9.8}$

$\mu = 0.327$

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28 days ago