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3 years ago

Which parts of a compound microscope magnify objects?

2 answers:

8 0

The parts of a compound microscope that magnify the image are the lenses. A compound microscope is called compound<span> because it has several lenses. The two main lenses are the ocular lens or eyepiece and the objective lenses.</span>

Yuri [45]3 years ago

7 0

The ocular lens or eyepiece lens

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Illustrate a body of mass 5.0kg pulled by horizontal force F . if the body accelerates 2.0ms² and experience a frictional force

Natasha2012 [34]

Explanation:

a. Net force is mass times acceleration (Newton's second law).

∑F = ma

∑F = (5.0 kg) (2.0 m/s²)

∑F = 10 N

b. The net force is the sum of the individual forces.

10 N = F − 5 N

F = 15 N

c. Friction force here is mgμ.

mgμ = 5 N

(5.0 kg) (10 m/s) μ = 5 N

μ = 0.1

3 0

2 years ago

3. A certain wire, 3 m long, stretches by 1.2 mm when under tension of 200 N. By how much does

nikitadnepr [17]

Answer:

The extension of the second wire is $e_2 = 0.0024 \ m = 2.4 mm$

Explanation:

From the question we are told that

The length of the wire is $L = 3 \ m$

The elongation of the wire is $e = 1.2mm = \frac{1.2}{1000} = 0.0012 m$

The tension is $F = 200 \ N$

The length of the second wire is $L_2 = 6 \ m$

Generally the Young's modulus(Y) of this material is

$Y = \frac{stress}{strain }$

Where $stress = \frac{F}{A}$

Where A is the area which is evaluated as

$A = \pi r^2$

and $strain = \frac{extention}{length} = \frac{e}{L}$

$Y = \frac{\frac{F}{\pi r^2 } }{ \frac{e}{L} }$

Since the wire are of the same material Young's modulus(Y) is constant

So we have

$\frac{F * L }{r^2 e} = \pi * Y = constant$

$F * L = constant * r^2 e$

Now the ration between the first and the second wire is

$\frac{F_1}{F_2} * \frac{L_1}{L_2} = \frac{r*2_1}{r^2} * \frac{e_1}{e_2}$

Since tension , radius are constant

We have

$\frac{L_1}{L_2} = \frac{e_1}{e_2}$

substituting values

$\frac{3}{6} = \frac{0.0012}{e_2}$

$0.5 e_2 = 0.0012$

$e_2 = \frac{ 0.0012 }{0.5}$

$e_2 = 0.0024 \ m = 2.4 mm$

3 0

3 years ago

State 2 chemical properties of metals

Ivahew [28]

Answer:

#they react with oxygen to form basic oxides.

#they react with water to produce hydrogen and oxides

#they react with acids to produce H+

3 0

2 years ago

Read 2 more answers

A construction worker drags a box across the floor. If the frictional force between the floor and the box is 12.2 newtons, how m

victus00 [196]

The work done is 140.3 J

Explanation:

The work done is defined as the amount of force required or used to move an object over a particular distance. Hence, work done is the product of force in Newton and the distance in which the object moved in meters.
Here, in the question, it is given that there is a frictional force acting on the box by the floor and the distance it moves. Hence, to get the answer to multiply frictional force and the distance. So, will get the work done as 140.3 J.

8 0

3 years ago

What is the area of the circle with a radius of 12.77m

Arada [10]

A=pi r^2
A=( 3.14159.......)(12.77)^2
A= 512.30m^2

8 0

3 years ago