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3 years ago

11

Find the separation of two points on the Moon’s surface that can just be resolved by the 200 in. (= 5.1 m) telescope at Mount Pa

lomar, assuming that this separation is determined by diffraction effects. The distance from Earth to the Moon is
3.8×105km
. Assume a wavelength of 550 nm for the light.

1 answer:

rewona [7]3 years ago

6 0

Answer:

The separation of the 2 points should be 50.0 meters.

Explanation:

According to Rayleigh's scattering criteria the angular separation between 2 points to be resolved equals

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

Applying the given values we get

$\theta =1.22\cdot \frac{550\times 10^{-9} }{5.1}=0.1316\times 10^{-6}rads$

thus the linear separation equals $L=\theta \times Distance$

Applying the given values we get

$L=0.1316\times 10^{-6} \times 3.8\times 10^{5}\times 10^{3}meters\\\\\therefore L=50.00metes$

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6 0

2 years ago

Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates

olasank [31]

2) 20.2 m/s

In the first 4.4 seconds of its motion, the blue car accelerates at a rate of

$a=4.6 m/s^2$

So its final velocity after these 4.4 seconds is

$v=u+at$

where

u = 0 is the initial velocity (the car starts from rest)

a is the acceleration

t is the time

Substituting t = 4.4 s, we find

$v=0+(4.6)(4.4)=20.2 m/s$

After this, the car continues at a constant speed for another 8.5 s, so it will keep this velocity until

$t=4.4 + 8.5 = 12.9 s$

Therefore, the velocity of the car 10.4 seconds after it starts will still be 20.2 m/s.

3) 216.2 m

The distance travelled by the car during the first 4.4 s of the motion is given by

$d_1 = ut_1 + \frac{1}{2}at_1^2$

where

u = 0 is the initial velocity

$t_1 = 4.4 s$ is the time

$a=4.6 m/s^2$ is the acceleration

Substituting,

$d_1 = 0 +\frac{1}{2}(4.6)(4.4)^2=44.5 m$

The car then continues for another 8.5 s at a constant speed, so the distance travelled in the second part is

$d_2 = vt_2$

where

$v_2 = 20.2 m/s$ is the new velocity

$t_2 = 8.5 s$ is the time

Substituting,

$d_2 = (20.2)(8.5)=171.7 m$

So the total distance travelled before the brakes are applied is

$d=44.5 m+171.7 m=216.2 m$

4) $-6.62 m/s^2$

We are told that the blue car comes to a spot at a distance of 247 meters from the start. Therefore, the distance travelled by the car while the brakes are applied is

$d_3 = 247 m -216.2 m=30.8 m$

We can find the acceleration of the car during this part by using the SUVAT equation:

$v_f^2 - v_i^2 = 2ad_3$

where

$v_f = 0$ is the final velocity (zero since the car comes to a stop)

$v_i = 20.2 m/s$ is the velocity of the car at the moment the brakes are applied

a is the acceleration

$d_3 = 30.8 m$

Solving for a, we find

$a=\frac{v_f^2 -v_i^2 }{2d}=\frac{0-(20.2)^2}{2(30.8)}=-6.62 m/s^2$

5 0

3 years ago

300kg of water are lifted 10m vertically in 5s show the work done in 30kj and that power is 6kw . Please help me

Mandarinka [93]

Answer:

6KW

Explanation:

The computation is shown below:

We know that

Work done= m ×g× h

Here

W= 300×10×10

= 30000 J

= 30 KJ

And

Power= Work done ÷time taken

P = 30000 ÷ 5

= 6000W

= 6KW

The above represent the answer

3 0

2 years ago

What is the wavelength of the wave in the image to the left?<br><br><br> Graph 1

cupoosta [38]

Answer:

There is no graph i can aswer it soww

Explanation:

4 0

3 years ago

What is the shortest possible time in which a bacterium to travel distance of 8.4cm across a Petri dish at a constant velocity o

Dimas [21]

Answer:

$\boxed{\sf Shortest \ possible \ time = 7 \ seconds}$

Given:

Distance travelled (s) = 8.4 cm

velocity (v) = 1.2 cm/s

To Find:

Shortest possible time (t) in which a bacterium travel a distance 8.4 cm across a Petri Dish

Explanation:

$\boxed{ \bold{\sf Time \ (t) = \frac{Distance \ travelled \ (s)}{Velocity \ (v)}}}$

Substituting values of Distance travelled (s) & Velocity (v) in the equation:

$\sf \implies t = \frac{8.4}{1.2}$

$\sf \implies t = \frac{7 \times \cancel{1.2}}{ \cancel{1.2}}$

$\sf \implies t = 7 \: s$

3 0

3 years ago