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slega [8]
1 year ago
9

When you changed from low to high power, how did the change affect the working distance of the lens?

Physics
1 answer:
Basile [38]1 year ago
4 0

The working distance gets shorter as the magnification gets bigger. In order to focus, the high-power objective lens must be significantly nearer to the specimen than the low-power lens. Magnification is negatively correlated with working distance.

Magnification change The magnification of a specimen is increased by switching from low power to high power. The magnification of an image is determined by multiplying the magnification of the objective lens by the magnification of the ocular lens, or eyepiece.

The geometry of the optical system connects the magnifying power, or how much the thing being observed seems expanded, and the field of view, or the size of the object that can be seen.

To know more about  working distance

brainly.com/question/13551539

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For a given amount of gas at a constant temperature, the volume of a gas varies inversely with its pressure is a statement of __
Ksivusya [100]

Answer:

d. Boyle's

Explanation:

Boyle's Law: States that the volume of a fixed mass of gas is inversely proportional proportional to its pressure, provided temperature remains constant.

Stating this mathematically. this implies that:

V∝1/P

V = k/P, Where k is the constant of proportionality

PV = k

P₁V₁ = P₂V₂

Where P₁ and P₂ are the initial and final pressure respectively, V₁ and V₂ are the the initial and final volume respectively.

Hence the right option is d. Boyle's

8 0
2 years ago
The British attempted to break the land-speed record and the sound barrier using a jet-powered car. They made their attempt in t
jolli1 [7]

Your answer is B.

The relationship between medium temperature and speed of sound is a direct relationship: when one factor increases, the other increases as is shown in graph B. The British would choose the the time of day which would give the lowest speed of sound, because this would be easiest to break. Graph B shows that the lowest speed of sound would occur with the lowest air temperature - in the morning.

4 0
3 years ago
Read 2 more answers
Pleaseeee Please help, I will love you forever and ever
matrenka [14]

Answer:

The answer to your question is

Explanation:

Data

mass = 0.5kg

T1 = 35

T2 = ?

Q = - 6.3 x 10⁴ J  = - 63000 J

Cp = 4184 J / kg°C

Formula

                        Q = mCp(T2 - T1)

                         T2 = T1 + Q/mCp    

Substitution

                       T2 = 35 - 63000/(0.5 x 4184)

                        T2 = 35 - 63000/2092

                        T2 = 35 - 30.1

                         T2 = 4.9 °C

6 0
2 years ago
What is the main determining factor in defining boundaries between layers of earth's atmosphere?
konstantin123 [22]
The main determining factor in defining boundaries between layers of earth's atmosphere would be temperature changes in these layers. Temperature is one essential property that varies in the atmosphere. Based from this variation, the atmosphere is divided into four major layers and further to three smaller layers -  troposphere, tropopause, the stratosphere,  stratopause, the mesosphere, mesopause, and the thermosphere.The troposphere is the layer that is nearest to the surface of the Earth. It is the part where humans, plants and animals survive. Also, it is the warmest layer of the atmosphere. And as we go higher the atmosphere, the temperature would drop making it much cooler.
5 0
3 years ago
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A 6.0 g marble is fired vertically upward using a spring gun. The spring must be compressed 9.4 cm if the marble is to just reac
RoseWind [281]

Answer:

a) \Delta U_{g} = 12.945\,J, b) \Delta U_{k} = 12.945\,J, c) k = 2930.059\,\frac{N}{m}

Explanation:

a) The change in the gravitational potential energy of the marble-Earth system is:

\Delta U_{g} = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (22\,m)

\Delta U_{g} = 12.945\,J

b) The change in the elastic potential energy of the spring is equal to the change in the gravitational potential energy, then:

\Delta U_{k} = 12.945\,J

c) The spring constant of the gun is:

\Delta U_{k} = \frac{1}{2} \cdot k \cdot x^{2}

k = \frac{2\cdot \Delta U_{k}}{x^{2}}

k = \frac{2\cdot (12.945\,J)}{(0.094\,m)^{2}}

k = 2930.059\,\frac{N}{m}

4 0
2 years ago
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