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2 years ago

which part of the microscope will be used first to adjust the focus when starting with the lowest power lens?

1 answer:

7 0

The first part of the microscope that should first be used to adjust the focus when starting with the lowest power lens would be the coarse adjustment knob.

There are two knobs in a typical light microscope with which objects on slides can be brought into focus:

Coarse adjustment knob
Fine adjustment knob

The 2 knobs are used to adjust the stage to either bring it up towards the objective lens or down away from them. The coarse adjustment knob, however, moves the stage a considerable distance with each turn. The fine adjustment knob, on the other hand, only moves the stage very little with each turn.

The lowest power lenses are often short. Hence, using the coarse adjustment knob is ideal in order to quickly bring objects on slides into focus.

The fine adjustment knob comes highly recommended at high objectives because high objectives lenses are usually long and using the coarse adjustment knob can lead to a breakage of the slide by the lens.

More on bringing objects into focus on a microscope can be found here: brainly.com/question/24319677

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2 years ago

Why does the arrow labeled F norm point upward?

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1 year ago

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A wire 6.90 m long with diameter of 2.15 mm has a resistance of 0.0320 Ω. Find the resistivity of the material of the wire. rho

spayn [35]

<h2>Answer:</h2>

1.68 x 10⁻⁸Ωm

<h2>Explanation:</h2>

The resistance (R) of a wire is related to its length(L), its material resistivity(ρ) and its crossectional area(A) as follows;

R = ρL/A ------------------------(i)

Where;

A = πd² / 4 [where d = diameter of the wire]

From the question;

L = 6.90m

d = 2.15mm = 0.00215m

R = 0.0320Ω

First calculate the crossectional area (A) of the wire as follows;

A = πd² / 4

[Take π = 3.142]

d = 0.00215m

∴ A = 3.142 x (0.00215)² / 4

∴ A = 0.000003631m²

Now, substitute the values of A, L, and R into equation (i) as follows;

R = ρL/A

0.0320 = ρ x 6.90 / 0.000003631

0.0320 = 1900302.95 x ρ

Solve for ρ;

=> ρ = 0.0320 / 1900302.95

=> ρ = 1.68 x 10⁻⁸Ωm

Therefore, the resistivity of the material of the wire is 1.68 x 10⁻⁸Ωm

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Power can be defined as?

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2 years ago

An electron is isolated from an atom and exists in vacuum. A group of scientists collectively state that they can remove part of

lesya692 [45]

D) Partial charge cannot be removed, because charge is a discrete quantity that may exist only at certain values

Explanation:

The electric charge of an object is a property of the object that is related to the ability of the object to experience/exert an electric force: if the object is electrically charge, then it is attracted or repelled by other electrically charged object.

The electric charge of an object depends on the amount of charged particles it has on it. In particular, the fundamental particles that carry electric charge are:

Protons: they carry electric charge of +e
Electrons: they carry electric charge of -e

Where "e" is the fundamental charge ( $e=1.6\cdot 10^{-19}C$ ). Therefore, one proton carry a charge of +e and one electron carry a charge of -e.

An electron is a fundamental particle: this means that it cannot be divided into smaller particles. This also means that it is not possible to remove part of the charge of the electron: in fact, it is said that electric charge exists only as discrete values, being a multiple of $e$ . Therefore, the correct statement is

D) Partial charge cannot be removed, because charge is a discrete quantity that may exist only at certain values

Learn more about particles:

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