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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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1 year ago

A tiny object carrying a charge of +44 μC and a second tiny charged object are initially very far apart. If it takes 21 J of wor

STatiana [176]

Answer:

The magnitude of the second charge is $\rm 1.062\times 10^{-7}\ C$ or $\rm 0.1062\ \mu C.$

Explanation:

The work done in bringing a charged particle from one point to another in the presence of some electric field is equal to the change in the electric potential energy of the charge in moving from one point to another.

The electric potential energy of some charge $q_o$ at a point in the electric field of another charge $q$ is given by the product of the amount of charge $q_o$ and electric potential at that point due to the charge $q$ .

$U = q_o\ V.$

The electric potential at that point is given by

$V = \dfrac{kq}{r}.$

where $k$ is the Coulomb's constant.

Therefore,

$U=q_o\ \dfrac{kq}{r}.$

Now, We have given two charges $q_1 = +44\ \mu C = +44\times 10^{-6}\ C$ and $q_2$ , whose value is to be found.

When the two charges are infinitely dar apart, the electric potential energy of the system is given by

$U_i = \dfrac{kq_1q_2}{\infty}=0.$

When the coordinates of position of the two charges are

$(x_1,\ y_1) = (1.00\ mm,\ 1.00\ mm).\\(x_2,\ y_2) = (1.00\ mm,\ 3.00\ mm).$

The distance between the two charges is given by

$r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(1.00-1.00)^2+(3.00-1.00)^2}=2.00\ mm = 2.00\times 10^{-3}\ m.$

The electric potential energy of the charges in this configuration is given by

$U_f = \dfrac{kq_1q_2}{r}\\=\dfrac{(8.99\times 10^9)\times (+44\times 10^{-6})\times q_2}{2.00\times 10^{-3}}\\=1.9778\times 10^8\times q_2.$

The change in the electric potential energy of the system is equal to the work done to bring the system from inifinitely far apart position to given configuration.

Therefore,

$W = U_f-U_i\\21=(1.9778\times 10^8\times q_2)-0\\\Rightarrow q_2 = \dfrac{21}{1.9778\times 10^8}\\=1.062\times 10^{-7}\ C\\=0.1062\times 10^{-6}\ C\\=0.1062\ \mu C.$

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3 years ago