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3 years ago

Why is wavelength the main limiting factor on limit of resolution in light microscopy?

1 answer:

3 0

The resolution of a microscope is the smallest resolvable distance between two objects. It is defined as the shortest distance between two points on a specimen and the observer can still distinguish them.
The wavelength is a determining and limiting factor in the degree of resolution afforded by the microscope. The relationship between the wavelength and the resolution is:
Shorter wavelengths yield higher resolution and visa versa.

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The spaceship Intergalactica lands on the surface of the uninhabited Pink Planet, which orbits a rather average star in the dist

Sidana [21]

Complete Question

The spaceship Intergalactica lands on the surface of the uninhabited Pink Planet, which orbits a rather average star in the distant Garbanzo Galaxy. A scouting party sets out to explore. The party's leader–a physicist, naturally–immediately makes a determination of the acceleration due to gravity on the Pink Planet's surface by means of a simple pendulum of length 1.08m. She sets the pendulum swinging, and her collaborators carefully count 101 complete cycles of oscillation during 2.00×102 s. What is the result? acceleration due to gravity:acceleration due to gravity: m/s2

Answer:

The acceleration due to gravity is $g = 167.2 \ m/s^2$

Explanation:

From the question we are told that

The length of the simple pendulum is $L = 1.081.08 \ m$

The number of cycles is $N = 101$

The time take is $t = 2.00 *10^{2 \ }s$

Generally the period of this oscillation is mathematically evaluated as

$T = \frac{N}{t }$

substituting values

$T = \frac{101}{2.0*10^2 }$

$T = 0.505 \ s$

The period of this oscillation is mathematically represented as

$T = 2 \pi \sqrt{\frac{l}{g} }$

making g the subject of the formula we have

$g = \frac{L}{[\frac{T}{2 \pi } ]^2 }$

$g = \frac{4 \pi ^2 L }{T^2 }$

Substituting values

$g = \frac{4 * 3.142 ^2 * 1.08 }{505.505^2 }$

$g = \frac{4 * 3.142 ^2 * 1.08 }{0.505^2 }$

$g = 167.2 \ m/s^2$

7 0

2 years ago

PROBLEM 5 (Problem 4-145 in 7th edition) Consider a well-insulated horizontal rigid cylinder that is divided into two compartmen

kari74 [83]

Answer:

The answer is " $\bold{83.8^{\circ} \ C}$ ".

Explanation:

Formula for calculating the mass in He:

$\to m = \frac{PV}{RT}\\$

$= \frac{500 \times 1}{ 2.0769 \times (40 + 273)}\\\\ = \frac{500 }{ 2.0769 \times 313}\\\\ = \frac{500 }{ 650.0697}\\\\= 0.76914 \ Kg$

Formula for calculating the mass in $N_2$ :

$\to m = \frac{PV}{RT}\\$

$= \frac{500 \times 1}{ 0.2968 \times (120+ 273)}\\\\ = \frac{500 }{ 0.2968 \times 393}\\\\ = \frac{500 }{ 116.6424}\\\\= 4.2866\ Kg$

by using the temperature balancing the equation:

$T' = \frac{mcT (He) + mcT ( N_2 )}{ mc (He) + mc ( N_2)}$

$= \frac{0.76914 \times 3.1156 \times 313 + 4.2866 \times 0.743 \times393}{ 0.76914 \times 3.1156 + 4.2866 \times 0.743} \\\\ = 357 \ \ K \approx 83.8^{\circ} \ C$

3 0

2 years ago

What is newton's third law second law first law ?

sergeinik [125]

Answer:

first law: an object remains in uniform motion except an external force has acted upon it eg a ball in stable motion doesn't move until one moves or kicks it

second law:the body acted upon by an external force gains a momentum which is directly proportional to the applied force and acts in the direction of the force

third law: to every action there is an equal and opposite reaction eg if u push someone the person moves backward away from you and not towards you

8 0

3 years ago

Read 2 more answers

A thin lens with a focal length of 6.0 cm is used as a simple magnifier by (a) what angular magnification is obtainable with the

Serhud [2]

Answer:

4.167

4.83871 cm

Explanation:

u = Object distance

v = Image distance = 25 cm

f = Focal length = 6 cm

Angular magnification is given by

$m=\frac{25}{f}\\\Rightarrow m=\frac{25}{6}\\\Rightarrow m=4.167$

The angular magnification of the lens is 4.167

Lens equation

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{6}=\frac{1}{u}+\frac{1}{-25}\\\Rightarrow \frac{1}{6}+\frac{1}{25}=\frac{1}{u}\\\Rightarrow \frac{1}{u}=\frac{31}{150}\\\Rightarrow u=4.83871\ cm$

The closest distance by which the object can be examined is 4.83871 cm

5 0

2 years ago

What is the exact distance from the Earth to the Sun?

notka56 [123]

94.5 million miles.

I hope this helps, please Brainliest me, and have a great night! :D

8 0

3 years ago

Read 2 more answers