Ask question

Ask question

All categories

4 years ago

5

The girl makes a microscope with a 3.0 cm focal length objective and a 5.0 cm eyepiece. The microscope tube length is 10 cm. Use

the simple formula to find the expected magnification of this microscope.
(A) 12
(B) 14
(C) 17
(D) 20
(E) 24

1 answer:

saul85 [17]4 years ago

3 0

To solve this problem we will use the concepts related to Magnification. Magnification is the process of enlarging the apparent size, not physical size, of something. This enlargement is quantified by a calculated number also called "magnification".

The overall magnification of microscope is

$M = \frac{Nl}{f_ef_0}$

Where

N = Near point

l = distance between the object lens and eye lens

$f_0$ = Focal length

$f_e$ = Focal of eyepiece

Given that the minimum distance at which the eye is able to focus is about 25cm we have that N = 25cm

Replacing,

$M = \frac{25*10}{3*5}$

$M = 16.67\approx 17\\$

Therefore the correct answer is C.

You might be interested in

What are the different type of insulators in a torch?

Naily [24]

I believe they are:

Pin Insulators
Suspension Insulators
Strain Insulators

If thats not enough theres also insulators commonly used of lower voltage which are:
Stay
Shackle

6 0

4 years ago

The rate at which speed changes is called

pochemuha

Answer:

Acceleration

Explanation:

I think so & hope it will help yuh!

8 0

3 years ago

A 3-m-high, 7-m-wide rectangular gate is hinged at the top edge and is restrained by a fixed ridge. Determine the hydrostatic fo

Shalnov [3]

Answer:

The Hydrostatic force is $F = 137.2 kN$

The location of pressure center is $Z = 1.333 \ m$

Explanation:

From the question we are told that

The height of the gate is $h = 3 \ m$

The weight of the gate is $w = 7 \ m$

The height of the water is $h_w = 2 \ m$

The density of water is $\rho_w = 1000 \ kg/m^3$

Note used $h_w$ for height of water and height of gate immersed by water since both have the same value

The area of the gate immersed in water is mathematically represented as

$A = h_w * w$

substituting values

$A = 2* 7$

$A = 14 \ m^2$

The hydrostatic force is mathematically represented as

$F = \rho_w * g * h_f * A$

Where

$h_f =h- h_w$

$h_f =3 -2$

$h_f = 1\ m$

So

$F = 1000 * 9.8 * 1 * 14$

$F = 137.2 kN$

The center of pressure is mathematically represented as

$Z = h_f + \frac{I_g}{h_f * A}$

Where $I_g$ is the moment of inertia of the gate which mathematically represented as

$I_g = \frac{w * h_w^2}{12}$

The $h_w$ is the height of gate immersed in water

$I_g = \frac{7 * 2^2 }{12}$

$I_g = 4.667\ kg m^2$

Thus

$Z = 1 + \frac{4.66}{1 * 14}$

$Z = 1.333 \ m$

3 0

3 years ago

Reasons<br> why using mirrors to generate electricity is not a common practice.

olganol [36]

A big part of the reason that mirrors are seldom if ever used to generate
electricity is the simple fact that there is no way to generate electricity using
mirrors. They are as useless for the purpose as smoke is, although there are
those who have used both items simultaneously to create the impression that
they have succeeded in that attempt.

7 0

3 years ago

Read 2 more answers

Mercury has a density of 13.56 g/mL. How many kilograms of mercury would you expect to fit in a cylindrical glass cup with a bot

vovangra [49]

Answer:

263.152kg

Explanation:

<em>The density of a substance is related to its mass and volume as follows;</em>

density = mass / volume

mass = density x volume -------------(i)

The substance in question here is <em>mercury </em>which has;

density = 13.56g/mL = 13.56g/cm³

Since the mercury is going to be put in the cylindrical glass, the volume of the cylindrical glass is going to be equal to the volume of the mercury that will be put.

And we know that the;

volume of a cylinder = πr²h

<em>Where;</em>

π = 3.142

r = bottom radius of the cylinder = 5.75inches

h = height of the cylinder = 0.950ft

<em>For uniformity, let's convert the radius and height of the cylinder to their corresponding values in cm</em>

r = 5.75 inches = 5.75 x 2.54 cm = 14.605cm

h = 0.950 ft = 0.950 x 30.48 cm = 28.956cm

<em>Therefore, the volume of the cylinder;</em>

v = 3.142 x (14.605cm)² x 28.956cm = 19406.5cm³

v = 19406.5cm³ [This is also the volume of the mercury necessary to fit the cylinder]

<em>Now the following value has been found;</em>

volume = 19406.5cm³

<em>Substitute the values of density and volume into equation (i) as follows;</em>

mass = 19406.5cm³ x 13.56g/cm³

mass = 263152.14g

<em>Convert the result to kg by dividing by 1000</em>

mass = 263.152kg

Therefore, 263.152kg kilograms of mercury would fit in the cylindrical glass.

3 0

4 years ago