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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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A baseball pitcher throws a ball horizontally at a speed of 34.0 m/s. A catcher is 18.6 m away from the pitcher. Find the magnit
Sidana [21]

To develop this problem, it is necessary to apply the concepts related to the description of the movement through the kinematic trajectory equations, which include displacement, velocity and acceleration.

The trajectory equation from the motion kinematic equations is given by

y = \frac{1}{2} at^2+v_0t+y_0

Where,

a = acceleration

t = time

v_0 = Initial velocity

y_0 = initial position

In addition to this we know that speed, speed is the change of position in relation to time. So

v = \frac{x}{t}

x = Displacement

t = time

With the data we have we can find the time as well

v = \frac{x}{t}

t = \frac{x}{v}

t = \frac{18.6}{34}

t = 0.547s

With the equation of motion and considering that we have no initial position, that the initial velocity is also zero then and that the acceleration is gravity,

y = \frac{1}{2} at^2+v_0t+y_0

y = \frac{1}{2} gt^2+0+0

y = \frac{1}{2} gt^2

y = \frac{1}{2} 9.8*0.547^2

y = 1.46m

Therefore the vertical distance that the ball drops as it moves from the pitcher to the catcher is 1.46m.

6 0
2 years ago
Which of the following statements about cultural variations is true?
viktelen [127]

Answer:

i think its D but i could be wrong im sorry if i am

Explanation:

edge 2020

7 0
2 years ago
Calculate the number of moles of water molecules in 12 dm' of water<br>vapour at STP.<br><br>​
Vinvika [58]

Answer:

22.4 \:  {dm}^{3}  \: are \: occupied \: by \: 1 \: mole \\ 12 \:  {dm}^{3}  \: will \: be \: occupied \: by \: (  \frac{12}{22.4} ) \: moles \\  = 0.536 \: moles

4 0
2 years ago
) A stone initially moving at 8.0 m/s on a level surface comes to rest due to friction after it travels 11 m. What is the coeffi
natali 33 [55]

Answer:

-0.3

Explanation:

F' = μmg ........... Equation 1

Where F' = Frictional force, μ = coefficient of kinetic friction, m = mass of the stone, g = acceleration due to gravity.

But,

F' = ma ............ Equation 2

Where a = acceleration of the stone.

Substitute equation 2 into equation 1

ma = μmg

dividing both side of the equation by m

a = μg

make μ the subject of the equation

μ = a/g............... Equation 3

From the equation of motion,

v² = u²+2as................. Equation 4

Where v and u are the final and the initial velocity respectively, s = distance.

Given: v = 0 m/s (to rest), u = 8.0 m/s, s = 11 m.

Substitute into equation 4

0² = 8² + 2×11×a

22a = -64

a = -64/22

a = -32/11 m/s² = -2.91 m/s²

substitute the values of a and g into equation 3

μ = -2.91/9.8

μ = -0.297

μ ≈ -0.3

4 0
3 years ago
A heat engine exhausts 8900 j of heat while performing 2800 j of useful work. part a what is the efficiency of this engine?
Anuta_ua [19.1K]
Given: Heat Qout means useful work = 2800 J

           Heat Qin = 8900 J

Required; Efficiency = ?

Formula: Efficiency = Qout/Qin = x 100%

                               = 2800 J/8900 J = 0.3146 X 100 %

               Efficiency = 31.46%


7 0
3 years ago
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