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wariber [46]
2 years ago
4

A microscope is focused on an amoeba. Part A When a 0.15-mm-thick cover glass (n=1.50) is placed over the amoeba, by how far mus

t the microscope objective be moved to bring the organism back into focus? Express your answer in millimeters. ΔsΔ s = nothing mm Request Answer Part B Must it be raised or lowered?
Physics
1 answer:
Shalnov [3]2 years ago
3 0

Answer:

\Delta s=0.05\ mm

<em>∵The apparent depth is less therefore we need to raise the objective away from the placed object.</em>

Explanation:

Given:

Real depth, d_r=0.15\ mm

refractive index of the medium, n=1.5

We know,

n=\frac{d_r}{d_a}

where:

d_a= apparent depth of the object

\therefore 1.5=\frac{0.15}{d_a}

d_a=0.1\ mm

<u>Now, the difference in the real and apparent depths:</u>

\Delta s=d_r-d_a

\Delta s=0.05\ mm

∵The apparent depth is less therefore we need to raise the objective away from the placed object.

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Crank

Answer:

false.

Explanation:

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Remember that the velocity is a vector, so it has a direction.

Then when she goes from the 1st end to the other, the velocity is positive

When she goes back, the velocity is negative

if both cases the magnitude of the velocity, the speed, is the same, then the average velocity is:

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Answer:

Explanation:

From the data it appears that A is the middle point between two charges.

First of all we shall calculate the field at point A .

Field due to charge -Q ( 6e⁻ ) at A

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We shall add the field  to get the resultant field  .

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