Ask question

Ask question

All categories

3 years ago

5

You are given two converging lenses to build a compound microscope. Lens A has focal length 0.50 cm and lens B has focal length

of 3.0 cm. Which one of the two lenses would you use for the objective

1 answer:

siniylev [52]3 years ago

6 0

Answer:

The lens to be used for the objective is lens A

Explanation:

The objective of a compound microscope

The focal length of the lens used for the objective = 1/(magnification obtained)

The focal length of most modern is equal to the tube length

The range of sizes for the focal length of a microscope is between 2 mm and 40 mm

Therefore, the appropriate lens to be used for the objective of the compound is lens A that has a focal length of 0.50 cm = 5 mm

You might be interested in

What would be some benefits of living on a planet with less surface gravity

zepelin [54]

You could jump high!

3 0

3 years ago

Read 2 more answers

What is the gravitational potential energy of a 3 kg ball kicked into the air at a height of 5 meters?

sladkih [1.3K]

formula for gravitational P.E =mgh

Solution:-mass=3kg height=5metre and gravity=9.8 or 10m/sec² so P.E=mgh , 3×9.8×5=147kgm²/sec²

7 0

2 years ago

An astronaut finds herself in a predicament in which she has become untethered from her shuttle. She figures that she could get

Blizzard [7]

In order to solve the problem, it is necessary to apply the concepts related to the conservation of momentum, especially when there is an impact or the throwing of an object.

The equation that defines the linear moment is given by

$mV_i = (m-m_O)V_f - m_OV_O$

where,

m=Total mass

$m_O =$ Mass of Object

$V_i =$ Velocity before throwing

$V_f =$ Final Velocity

$V_O =$ Velocity of Object

Our values are:

$m_1=5.3kgm_2=7.9kg\\m_3=10.5kg\\m_A=75kg\\m_{Total}=m=98.7Kg$

Solving to find the final speed, after throwing the object we have

$V_f=\frac{mV_0+m_TV_O}{m-m_O}$

We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.

That way during each section the equations should be modified depending on the previous one, let's start:

A) $5.3Kg\rightarrow 15m/s$

$V_{f1}=\frac{mV_0+m_TV_O}{m-m_O}$

$V_{f1}=\frac{(98.7)*0+5.3*15}{98.7-5.3}$

$V_{f1}=0.8511m/s$

B) $7.9Kg\rightarrow 11.2m/s$

$V_{f2}=\frac{mV_{f1}+m_TV_O}{m-m_O}$

$V_{f2}=\frac{(98.7)(0.8511)+(7.9)(11.2)}{98.7-5.3-7.9}$

$V_{f2} = 2.0173m/s$

C) $10.5Kg\rightarrow 7m/s$

$V_{f3}=\frac{mV_{f2}+m_TV_O}{m-m_O}$

$V_{f3}=\frac{(98.7)(2.0173)+(10.5)(7)}{98.7-5.3-7.9-10.5}$

$V_{f3} = 3.63478m/s$

Therefore the final velocity of astronaut is 3.63m/s

7 0

3 years ago

Soloha48 [4]

I think it’s C b/c it works for me

3 0

2 years ago

A certain field line diagram illustrates the electric field due to three particles that carry charges 5.0 μC, -3.0 μC, and -2.0

4vir4ik [10]

Answer:

6

Explanation:

Number of lines emanate from + 5 micro coulomb is 15 .

They terminates at negative charges that means at - 3 micro coulomb and - 2 micro Coulomb.

the electric field lines terminates at - 3 micro Coulomb and - 2 micro Coulomb is in the ratio of 3 : 2.

So the lines terminating at - 3 micro coulomb

= $\frac{3}{5}\times 15 = 9$

So the lines terminating at - 2 micro coulomb

= $\frac{2}{5}\times 15 = 6$

So, the number of filed lines terminates at - 2 micro Coulomb are 6.

4 0

3 years ago