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ICE Princess25 [194]
3 years ago
11

Suppose you want a telescope that would allow you to see distinguishing features as small as 3.5 km on the Moon some 384,000 km

away. Assume an average wavelength of 550 nm for the light received.Required:What is the minimum diameter mirror on a telescope?
Physics
1 answer:
spayn [35]3 years ago
7 0

Explanation:

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

And, from equation ( 2 ), we get

\theta=\frac{x}{d}

Thus,

\frac{x}{d} &=1.22 \frac{\lambda}{D}

D &=1.22 \frac{\lambda d}{x}

=1.22 \frac{550 \times 10^{-9} 3.84 \times 10^{8}}{5 \times 10^{3}}

=0.0515 \mathrm{m}

Thus, the diameter of the telescope's mirror that would allow us to see details as small as is

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ivolga24 [154]

Answer:

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3 0
3 years ago
What would happen to the wavelength if the frequency was doubled?
ira [324]

Answer:

λ = hv

If frequency is doubled :

λ = h × 2v

λ = 2hv

Thus wavelength is doubled

7 0
3 years ago
A solid sphere of weight 42.0 N rolls up an incline at an angle of 36.0°. At the bottom of the incline the center of mass of the
Alecsey [184]

Answer:

Part a)

KE = 77.95 J

Part b)

L = 3.16 m

Part c)

distance L is independent of the mass of the sphere

Explanation:

Part a)

As we know that rotational kinetic energy of the sphere is given as

KE = \frac{1}{2}I\omega_2 + \frac{1}{2}mv^2

so we will have

KE = \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2 + \frac{1}{2}mv^2

so we will have

KE = \frac{1}{5} mv^2 + \frac{1}{2}mv^2

KE = \frac{7}{10} mv^2

KE = \frac{7}{10}(\frac{42}{9.81})(5.10^2)

KE = 77.95 J

Part b)

By mechanical energy conservation law we know that

Work done against gravity = initial kinetic energy of the sphere

So we will have

mgLsin\theta = KE

\frac{42}{9.81}(9.81)L sin36 = 77.95

L = 3.16 m

Part c)

by equation of energy conservation we know that

\frac{7}{10}mv^2 = mgL sin\theta

so here we can see that distance L is independent of the mass of the sphere

7 0
3 years ago
If you swing an object on the end of a string around a circle, the string pulls on the object to keep it moving in a circle. Wha
emmainna [20.7K]

Answer:

B

Explanation:

4 0
2 years ago
A baseball hit just above the ground leaves the bat 27 m/s at 45° above the horizontal. A) How far away does the ball strike the
Sedbober [7]

Answer:

A) The ball hits the ground 74.45 m far from the hitting position.

B) Maximum height of the ball = 18.57 m

Explanation:

There are two types of motion in this horizontal and vertical motion.

We have velocity = 27 m/s at 45° above the horizontal

Horizontal velocity = 27cos45 = 19.09 m/s

Vertical velocity = 27sin45 = 19.09 m/s

Time to reach maximum height,

           v = u + at

           0 = 19.09 - 9.81 t

            t = 1.95 s

So total time of flight = 2 x 1.95 = 3.90 s

A) So the ball travels at 19.09 m/s for 3.90 seconds.

     Horizontal distance traveled = 19.09 x 3.90 = 74.45 m

     So the ball hits the ground 74.45 m far from the hitting position.

B) We have vertical displacement

              S = ut + 0.5 at²

              H = 19.09 x 1.95 - 0.5 x 9.81 x 1.95² = 18.57 m

    Maximum height of the ball = 18.57 m

6 0
2 years ago
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