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DaniilM [7]
2 years ago
8

A(n) ________ microscope keeps an object in focus when the objective lens is changed.

Physics
1 answer:
Anvisha [2.4K]2 years ago
6 0

Parfocal is the term used to describe a microscope that maintains focus when the objective lenses are replaced.

<h3>What is the name of the objective lens ?</h3>

For observing minute features within a specimen sample, a high-powered objective lens, often known as a "high dry" lens, is perfect. You can see a very detailed image of the specimen on your slide thanks to the 400x total magnification that a high-power objective lens and a 10x eyepiece provide.

The four objective lenses on your microscope are for scanning (4x), low (10x), high (40x), and oil immersion (100x).

The first-stage lens used to create a picture from electrons leaving the specimen is referred to as the "objective lens." The objective lens is the most crucial component of the imaging system since the quality of the images is determined by how well it performs (resolution, contrast, etc.,).

To learn more than objective lens , visit

brainly.com/question/17307577

#SPJ4

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PLEASE I NEED HELP
777dan777 [17]

Answer:

K= 95.4 J

Explanation:

For this exercise we must use the conservation of mechanical energy.

We set a reference system on the floor.

Starting point. Higher

          Em₀ = U = m g h

Final point. Just before taking the floor

           Em_f = K = ½ m v²

energy is conserved because there is no friction

         Em₀ = Em_f

         mg h = K

         

The height is

  1.         h = y -y₀

                 h = 0- y₀

let's calculate

       K =  3.23 (-9.81)  (-3.01)          

       K= 95.4 J

5 0
3 years ago
What did Blaise Pascal discover, and how did he impact Physics?
yan [13]

Answer:

Well it really depends because he found out about a lot of things in a lot of scientific fields actually. He is known for contributing Pascal's triangle and probability theory. He also invented an early digital calculator and a roulette machine. In the field of physics, Blaise contributed to the study of atmospheric pressure by discovering that vacuums are real and exist in the real world.

6 0
3 years ago
How far from a converging lens with a focal length of 16 cm should an object be placed to produce a real image which is the same
Feliz [49]

Answer:

32 cm

Explanation:

f = focal length of the converging lens = 16 cm

Since the lens produce the image with same size as object, magnification is given as

m = magnification = - 1

p = distance of the object from the lens

q = distance of the image from the lens

magnification is given as

m = - q/p

- 1 = - q/p

q = p                                    eq-1

Using the lens equation, we get

1/p + 1/q = 1/f

using eq-1

1/p + 1/p = 1/16

p = 32 cm

4 0
3 years ago
True or false if an object has energy it must be moving
Drupady [299]

Answer:

Explanation:

If an object is moving, it is said to have kinetic energy (KE). Potential energy (PE) is energy that is "stored" because of the position and/or arrangement of the object. The classic example of potential energy is to pick up a brick. When it's on the ground, the brick had a certain amount of energy.

3 0
3 years ago
Read 2 more answers
. (a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110
7nadin3 [17]

Answer:

(a) the high of a hill that car can coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h is 47.6 m

(b) thermal energy was generated by friction is 1.88 x 10^{5} J

(C) the average force of friction if the hill has a slope 2.5º above the horizontal is 373 N

Explanation:

given information:

m = 750 kg

initial velocity, v_{0} = 110 km/h = 110 x 1000/3600 = 30.6 m/s\frac{30.6^{2} }{2x9.8}

initial height, h_{0} = 22 m

slope, θ = 2.5°

(a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h?

according to conservation-energy

EP = EK

mgh = \frac{1}{2} mv_{0} ^{2}

gh = \frac{1}{2} v_{0} ^{2}

h = \frac{v_{0} ^{2} }{2g}

  = 47.6 m

(b) If, in actuality, a 750-kg car with an initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its starting point, how much thermal energy was generated by friction?

thermal energy = mgΔh

                         = mg (h - h_{0})

                         = 750 x 9.8 x (47.6 - 22)

                         = 188160 Joule

                         = 1.88 x 10^{5} J

(c) What is the average force of friction if the hill has a slope 2.5º above the horizontal?

f d  = mgΔh

f = mgΔh / d,

where h = d sin θ, d = h/sinθ

therefore

f = (mgΔh) / (h/sinθ)

 = 1.88 x 10^{5}/(22/sin 2.5°)

 = 373 N

8 0
3 years ago
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