Ask question

Ask question

All categories

2 years ago

11

According to the rayleigh criterion, what is the shortest object we could resolve at the 25.0 cm near point with light of wavele

ngth 550 nm ?

1 answer:

Blizzard [7]2 years ago

7 0

Answer:

$6.71 *10^{-5}$ rad

Explanation:

∅ = $\frac{1.22*wavelength}{D}$ = $\frac{1.22*550 * 10^{-9} }{25 * 10^{-2} }$

∅ = $6.71 *10^{-5}$ rad

The minimum resolvable angle = $6.71 *10^{-5}$ rad

You might be interested in

When you step on the accelerator to increase the speed of your car, the force that accelerates the car is: A. the force of your

Dmitry [639]

Answer:

B. the force of friction of the road on the tires

Explanation:

Unless the car engine is like jet engine, the main force that accelerates the car forward is the force of friction of the road on the tires, which is ultimately driven by the force of engine on the tires shaft. As the engine, and the shaft are part of the system, their interaction is internal. According to Newton laws of motion, the acceleration needs external force, in this case it's the friction of the road on the tires.

6 0

2 years ago

What is the GPE of a 15,000 kg airplane sitting on the ground?

Vesna [10]

Answer:

C, it is not moving

it has no potential

7 0

2 years ago

How many miles away is the sun to the earth?

geniusboy [140]

<span>92.96 million mi..........</span>

7 0

3 years ago

Read 2 more answers

A 3 kg mass object is pushed 0.6 m into a spring with spring constant 210 N/m on a frictionless horizontal surface. Upon release

Svetradugi [14.3K]

Answer with Explanation:

We are given that

Mass of spring,m=3 kg

Distance moved by object,d=0.6 m

Spring constant,k=210N/m

Height,h=1.5 m

a.Work done to compress the spring initially= $\frac{1}{2}kx^2=\frac{1}{2}(210)(0.6)^2=37.8J$

b.

By conservation law of energy

Initial energy of spring=Kinetic energy of object

$37.8=\frac{1}{2}(3)v^2$

$v^2=\frac{37.8\times 2}{3}$

$v=\sqrt{\frac{37.8\times 2}{3}}$

v=5.02 m/s

c.Work done by friction on the incline, $w_{friction}=P.E-spring \;energy$

$W_{friction}=3\times 9.8\times 1.5-37.8=6.3 J$

8 0

2 years ago

a volleyball is hit upward with an initial velocity of 7.5 m/s. calculate the displacement of the volleyball when its final velo

Luden [163]

Answer:

The displacement of the volleyball is 2.62 m

Explanation:

Given;

initial velocity of the volleyball, u = 7.5 m/s

final velocity of the volleyball, v = 2.2 m/s

displacement of the volleyball, d = ?

Apply the following kinematic equation;

v² = u² - 2gd

2gd = u² - v²

$d = \frac{u^{2}-v^{2} }{2g}\\\\d = \frac{7.5^{2}-2.2^{2} }{2*9.8}\\\\d = 2.62 \ m$

Therefore, the displacement of the volleyball is 2.62 m

7 0

2 years ago