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3 years ago

An instrument for observing very small objects is

1 answer:

6 0

Microscope
A microscope shines a light through a slide with the sample of the object on and you look through a lens that allows you to magnify it by a very significant amount depending on what kind of microscope you have (a powerful electron microscope can magnify to the atomic level).

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The ceiling of your lecture hall is probably covered with acoustic tile, which has small holes separated by about 5.9 mm. Using

timurjin [86]

Answer:

45.88297 m

Violet

Explanation:

x = Gap between holes = 5.9 mm

$\lambda$ = Wavelength = 527 nm

D = Diameter of eye = 5 mm

L= Distance of observer from holes

From Rayleigh criteria we have the relation

$\frac{x}{L}=1.22\frac{\lambda}{D}\\\Rightarrow L=\frac{xD}{1.22\lambda}\\\Rightarrow L=\frac{5.9\times 10^{-3}\times 5\times 10^{-3}}{1.22\times 527\times 10^{-9}}\\\Rightarrow L=45.88297\ m$

A person could be 45.88297 m from the tile and still resolve the holes

Resolving them better means increasing the distance between the observer and the holes. It can be seen here that the distance is inversely proportional to the wavelength. Violet has a lower wavelength than red so, violet light would resolve the holes better.

5 0

2 years ago

A man fires a silver bullet of mass 2g with a velocity of200m/sec into wall. What is the temperature changeof the bullet? Note:

Sphinxa [80]

F=nmv

where;

n=no. of bullets = 1

m=mass of bullets=2g *10^-3

V=velocity of bullets200m/sec

F=1

loss in Kinetic energy=gain in heat energy

1/2MV^2=MS∆t

let M council M

=1/2V^2=S∆t

M=2g

K.E=MV^2/2

=(2*10^-3)(200)^2/2

2 councils 2

2*10^-3*4*10/2

K.E=40Js

H=mv∆t

(40/4.2)

40Js=40/4.2=mc∆t

40/4.2=2*0.03*∆t

=158.73°C

7 0

2 years ago

A wind turbine is rotating counterclockwise at 0.626 rev/s and slows to a stop in 12.9 s. Its blades are 17.9 m in length. What

iogann1982 [59]

Answer:

276.5 m/s^2

Explanation:

The initial angular velocity of the turbine is

$\omega=0.626 rev/s \cdot 2\pi rad/rev =3.93 rad/s$

The length of the blade is

r = 17.9 m

So the centripetal acceleration is given by

$a=\omega^2 r$

At the instant t = 0,

$\omega=3.93 rad/s$

So the centripetal acceleration of the tip of the blades is

$a=(3.93 rad/s)^2 (17.9 m)=276.5 m/s^2$

5 0

2 years ago

The bell rings and a physics student heads to class. They stop to talk to a few friends. They slow

grin007 [14]

Answer:

420m

Explanation:

Given parameters:

Time = 5minutes

Average speed = 1.4m/s

Unknown:

Distance covered = ?

Solution:

Speed is the rate of change of distance with time.

Mathematically;

Speed = $\frac{distance}{time}$

Distance = speed x time

Insert the parameters and solve;

Convert the time to seconds;

1 minute = 60s

5 minute = 5 x 60 = 300s

So,

Insert the parameters and find the distance;

Distance = 300 x 1.4 = 420m

3 0

2 years ago

A car is traveling at 108 km/h, stuck behind a slower car. Finally the road is clear and the car pulls over to make a pass. The

mezya [45]

Answer:

The average acceleration of the car is 2.143 meters per square second.

Explanation:

Let assume that car accelerates uniformly, in that case, we can obtain the value of acceleration by using the following equation of motion:

$v = v_{o}+a\cdot t$

Where:

$v_{o}$ - Initial velocity, measured in meters per second.

$v$ - Final velocity, measured in meters per second.

$a$ - Acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

Now, we clear acceleration within expression:

$a = \frac{v-v_{o}}{t}$

Initial and final velocities are now converted from kilometers per hour into meters per second:

$v_{o} = \left(108\,\frac{km}{h} \right)\cdot \left(1000\,\frac{m}{km} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s} \right)$

$v_{o} = 30\,\frac{m}{s}$

$v = \left(135\,\frac{km}{h} \right)\cdot \left(1000\,\frac{m}{km} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s} \right)$

$v = 37.5\,\frac{m}{s}$

If we know that $t = 3.5\,s$ , then, the average acceleration of the car is:

$a = \frac{37.5\,\frac{m}{s}-30\,\frac{m}{s} }{3.5\,s}$

$a = 2.143\,\frac{m}{s^{2}}$

The average acceleration of the car is 2.143 meters per square second.

8 0

2 years ago