Complete the calculations for total magnification produced by various combinations of the eyepiece and objective lenses. You may assume that the magnification for the eyepiece is 10X for each question. 1. When the scanning (4X) objective is used the total magnification will be:
2. When the low power (10X) objective is used the total magnification will be:
3. When the high power (40X) objective is used the total magnification will be:
4. When the oil immersion (100X) objective is used the total magnification will be:_

Answers

Answer 1

Answer:

Answer:

a) m_ttoal = 40x, b) m_total = 100X, c) m_total = 400X,

d) m_total = 1000 X

Explanation:

La magnificación o aumentos es el incremento de del tamaño de la imagen con respecto al tamaño original del objeto, en la mayoria del os sistema optico la magnificacion total es el producoto de la magnificación del objetivo por la magnificación del ocular

m_total = m_ objetivo * m=ocular

apliquemos esto a nuestro caso

1) m_total = 4 x * 10 x

m_ttoal = 40x

2) m_total = 10X * 10X

m_total = 100X

3)mtotal = 40X * 10X

m_total = 400X

4) m _totla = 100x * 10 X

m_total = 1000 X

en este ultimo caso para magnificación grandes es decalcificar el objeto

Answer 2

Answer:

The total magnification produced by different combinations of eyepiece and objective lenses in a microscope.

1. When the scanning (4X) objective is used, the total magnification will be 40X because the eyepiece magnification is 10X and the objective magnification is 4X.

2. When the low power (10X) objective is used, the total magnification will be 100X because the eyepiece magnification is 10X and the objective magnification is 10X.

3. When the high power (40X) objective is used, the total magnification will be 400X because the eyepiece magnification is 10X and the objective magnification is 40X.

4. When the oil immersion (100X) objective is used, the total magnification will be 1000X because the eyepiece magnification is 10X and the objective magnification is 100X.

Learn more about Microscope magnifications here:

brainly.com/question/36268796

#SPJ3

Related Questions

Air is a good conductor of heat. Please select the best answer from the choices provided T F
(a) How many kilograms of water must evaporate from a 60.0-kg woman to lower their body temperature by 0.750ºC?
Of the four most important pathways by which stress affects health, the first one to occuris usually related to physiology
Two non-conducting slabs of infinite area are given a charge-per-unit area of σ = -16 C/m^2 and σb =+6.0 C/m^2 respectively. A third slab, made of metal, is placed between the first two plates. The charge density σm on the metal slab is 0 (i.e., the slab is uncharged).Required:Find the magnitude and direction of the electric field.
An advantage of J.J. Thomson's Plum Pudding Model was that it _____. A. was a much less expensive way to study atoms B. simplified the calculations necessary to describe an atomC. clearly explained where electrons were located in an atomD. is much less expensive to bake a plum pudding than to look at an atom

If a single circular loop of wire carries a current of 61 A and produces a magnetic field at its center with a magnitude of 1.70 10-4 T, determine the radius of the loop.

Answers

Answer:

The radius is $R = 0.22 5 \ m$

Explanation:

From the question we are told that

The current is $I = 61 \ A$

The magnetic field is $B = 1.70 *10^(-4) \ T$

Generally the magnetic field produced by a current carrying conductor is mathematically represented as

$B = (\mu_o * I)/(2 * R )$

=> $R = (\mu_o * I )/( 2 * B )$

Here $\mu_o$ is the permeability of free space with value $\mu_o = 4\pi * 10^(-7) N/A^2$

=> $R = ( 4\pi * 10^(-7) * 61 )/( 2 * 1.70 *10^(-4) )$

=> $R = 0.22 5 \ m$

Find the intensity III of the sound waves produced by four 60-WW speakers as heard by the driver. Assume that the driver is located 1.0 mm from each of the two front speakers and 1.5 mm from each of the two rear speakers.

Answers

Given that,

Power = 60 W

Distance = 1.0 m

Distance between speakers = 1.5 m

We need to calculate the intensity

Using formula of intensity

$I_(1)=(P)/(A)$

$I_(1)=(P)/(4\pi r^2)$

Put the value into the formula

$I_(1)=(60)/(4\pi*(1.0)^2)$

$I_(1)=4.77\ W/m^2$

We need to calculate the intensity

Using formula of intensity

$I_(2)=(P)/(A)$

$I_(2)=(P)/(4\pi r^2)$

Put the value into the formula

$I_(1)=(60)/(4\pi*(1.5)^2)$

$I_(1)=2.12\ W/m^2$

We need to calculate the intensity of the sound waves produced by four speakers

Using formula for intensity

$I'=(I_(1)+I_(2))*2$

Put the value into the formula

$I'=(4.77+2.12)*2$

$I'=13.78\ W/m^2$

Hence, The intensity of the sound waves produced by four speakers is 13.78 W/m².

If Earth were completely blanketed with clouds and we couldn’t see the sky, could we learn about the realm beyond the clouds? What forms of radiation might penetrate the clouds and reach the ground?

Answers

The definition of waves that propagate through electric fields is called electromagnetic waves. The earth, despite being covered with clouds, can be 'affected' because waves such as sunlight or the moon have the ability to penetrate and be visible to the inhabitants of the earth. Microwaves and radio waves would be less affected by the clouds that cover the Earth.

Through these waves, you can know that there is beyond the clouds.

Ultraviolet light, microwaves and radio waves are the radiations that penetrate through the clouds and reach the Earth's surface.

Therefore, the answer is Yes, ultraviolet light, microwaves and radio waves are the forms of radiation that penetrate and reach the ground.

Final answer:

It is indeed possible to learn about the universe beyond the clouds due to other non-visual forms of radiation, mainly radio waves and gamma rays, which can penetrate through the clouds and reach the earth's surface.

Explanation:

Yes, even if Earth were completely blanketed with clouds and we could not see the sky, we could still learn about the universe beyond the clouds. This is because, in addition to visible light which would be blocked by the clouds, the universe also emits various other forms of radiation that can penetrate the clouds and reach the ground.

Two major types of radiation that could penetrate the dense clouds are radio waves and gamma rays. Radio waves are a form of electromagnetic radiation used in many areas of science and technology, while gamma rays are highly energetic forms of radiation and are used in fields such as astronomy to get valuable information about distant celestial bodies.

Learn more about radiation here:

brainly.com/question/4075566

#SPJ3

Patricia places 1500g of fruit on a scale compressing it by 0.02m. What is the force constant of the spring on the scale. Hint: you need 2 equations to solve this.

Answers

Answer:

The force constant of the spring is 735 N/m.

Explanation:

It is given that,

Mass of fruit, m = 1500 g = 1.5 kg

Compression in the scale, x = 0.02 m

We need to find the force constant of the spring on the scale. The force acting on the scale is given by using Hooke's law. So,

$F=-kx$

Also, F = mg

$mg=kx$

k is force constant

$k=(mg)/(x)\n\nk=(1.5* 9.8)/(0.02)\n\nk=735\ N/m$

So, the force constant of the spring is 735 N/m.

The electric field at the distance of 3.5 meters from an infinite wall of charges is 125 N/C. What is the magnitude of the electric field 1.5 meters from the wall?

Answers

Explanation:

It is given that,

Distance, r = 3.5 m

Electric field due to an infinite wall of charges, E = 125 N/C

We need to find the electric field 1.5 meters from the wall, r' = 1.5 m. Let it is equal to E'. For an infinite wall of charge the electric field is given by :

$E=(\lambda)/(2\pi \epsilon_o r)$

It is clear that the electric field is inversely proportional to the distance. So,

$(E)/(E')=(r')/(r)$

$E'=(Er)/(r')$

$E'=(125* 3.5)/(1.5)$

E' = 291.67 N/C

So, the magnitude of the electric field 1.5 meters from the wall is 291.67 N/C. Hence, this is the required solution.

Net force is the sum of all the forces acting on an object. If a spring balance pulls on a body with a force of 10 N, and friction acts on the body in the opposite direction with a force of 1 N, the net force would be 9 N in the direction of the spring balance (10 N – 1 N = 9 N).What is the net force acting on the object when the spring balance pulls the rope with a force of 25 N and friction acts on the body with a force of 20N?

Answers

Answer:

Explanation:

(25 N - 20 N = 5 N)

Other Questions

A spectroscope:measures light from distant objects makes object far away look closer receives radio signals from objects in space

When a body falls freely under gravity, then the work done by the gravity is ___________

A spectroscope breaks light up into its colors, allowing scientists to analyze light from the solar system and universe. By studying the spectral line patterns, widths, strengths and positions, scientists can determine the speed, position, and _____ of celestial bodies.A) age B) origin C) rotation D) temperature

Find an expression for the center of mass of a solid hemisphere, given as the distance R from the center of the flat part of the hemisphere. Express your answer in terms of R. Express the coefficients using three significant figures.