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JulsSmile [24]
3 years ago
5

A telescope can be used to enlarge the diameter of a laser beam and limit diffraction spreading. The laser beam is sent through

the telescope in opposite the normal direction and can then be projected onto a satellite or the Moon. If this is done with the Hale telescope, producing a 5.08 m diameter beam of 613 nm light, what is the minimum angular spread of the beam?
Physics
1 answer:
Novay_Z [31]3 years ago
5 0

Answer:

Angular spread, \theta=1.472\times 10^{-7}\ rad

Explanation:

It is given that,

Wavelength of the light, \lambda=613\ nm=613\times 10^{-9}\ m

Diameter of the telescope, D = 5.08 m

The minimum angular spread is given by :

\theta=\dfrac{1.22\lambda}{D}

\theta=\dfrac{1.22\times 613\times 10^{-9}}{5.08}

\theta=1.472\times 10^{-7}\ rad

So, the minimum angular spread of the beam is 1.472\times 10^{-7}\ radian. Hence, this is the required solution.

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What is the kinetic energy of a 120-cm thin uniform rod with a mass of 450 g that is rotating about its center at 3.60 rad/s?
goldfiish [28.3K]

Answer:

1.05 J.

Explanation:

Kinetic Energy: This is the energy possessed by a body due to its motion. The S.I unit of kinetic energy is Joules (J). The formula of kinetic energy is given as

Ek = 1/2mv²................. Equation 1

Where Ek = kinetic energy, m = mass of the uniform rod, v = liner velocity of the rod.

But,

v = αr .......................... Equation 2

Where α = angular velocity of the rod, r = radius of the circle.

Given: α = 3.6 red/s, r = 120/2 = 60 cm = 0.6 m.

Substitute into equation 2

v = 3.6(0.6)

v = 2.16 m/s.

Also given: m = 450 g = 0.45 kg.

Substitute into equation 1

Ek = 1/2(0.45)(2.16²)

Ek = 1.05 J.

4 0
2 years ago
A 782.10 kg car is brought from 7.60 m/s to 3.61 m/s over a time period of 4.23 seconds. What is the average force the car exper
alexandr402 [8]

Answer:

–735.17 N

The negative sign indicate that the force is acting in opposition direction to the car.

Explanation:

The following data were obtained from the question:

Mass (m) of car = 782.10 kg

Initial velocity (u) = 7.60 m/s

Final velocity (v) = 3.61 m/s

Time (t) = 4.23 s

Force (F) =?

Next, we shall determine the acceleration of the car. This can be obtained as follow:

Initial velocity (u) = 7.60 m/s

Final velocity (v) = 3.61 m/s

Time (t) = 4.23 s

Acceleration (a) =?

a = (v – u) / t

a = (3.61 – 7.60) / 4.23

a = –3.99 / 4.23

a = –0.94 m/s²

Finally, we shall determine the force experienced by the car as shown below:

Mass (m) of car = 782.10 kg

Acceleration (a) = –0.94 m/s²

Force (F) =?

F = ma

F = 782.10 × –0.94

F = –735.17 N

The negative sign indicate that the force is acting in opposition direction to the car.

4 0
2 years ago
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