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3 years ago

Write about any 8 areas of field where physics is applied. illustrate with examples.

Physics

1 answer:

Ulleksa [173]3 years ago

4 0

1. In construction. When constructing roads, building

2. In Engineering, when engineering we use physics such as calculations.

3. At home

4. In industries, industries use chemicals and so on.

5. In tailoring, we use the knowledge of physics like when sowing.

6. In carpentry, we use calculations.

7. In Hospitals, we use physics we balancing

8. In Sites, we use energy.

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As the wavelength of an electromagnetic wave ___ the frequency of the wave ________.

adoni [48]

Answer:

As the wavelength of an electromagnetic wave _decrease__ the frequency of the wave _increase_______.

Explanation:

What is the relationship between frequency and wavelength?

Wavelength and frequency of light are closely related. The higher the frequency, the shorter the wavelength. Because all light waves move through a vacuum at the same speed, the number of wave crests passing by a given point in one second depends on the wavelength.

That number, also known as the frequency, will be larger for a short-wavelength wave than for a long-wavelength wave. The equation that relates wavelength and frequency is:

V= fλ

where v= velocity

f= frequency

λ = wavelength

⇒ f = v/λ

also f ∝ 1/λ

For electromagnetic radiation, the speed is equal to the speed of light, c, and the equation becomes:

C= fλ

where c= Speed of light

f= frequency

λ = wavelength

⇒ f = v/λ

also f ∝ 1/λ

8 0

4 years ago

A screen is placed 1.60 m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40 c

Lunna [17]

Answer:

distance between the two second-order minima is 2.8 cm

Explanation:

Given data

distance = 1.60 m

central maximum = 1.40 cm

first-order diffraction minima = 1.40 cm

to find out

distance between the two second-order minima

solution

we know that fringe width = first-order diffraction minima /2

fringe width = 1.40 /2 = 0.7 cm

and

we know fringe width of first order we calculate slit d

β1 = m1λD/d

d = m1λD/β1

and

fringe width of second order

β2 = m2λD/d

β2 = m2λD / ( m1λD/β1 )

β2 = ( m2 / m1 ) β1

we know the two first-order diffraction minima are separated by 1.40 cm

y = 2β2 = 2 ( m2 / m1 ) β1

put here value

y = 2 ( 2 / 1 ) 0.7

y = 2.8 cm

so distance between the two second-order minima is 2.8 cm

6 0

3 years ago

In 8.5 s a fisherman winds 2.4 m of fishing line onto a reel whose radius is 3.0 cm (assumed to be constant as an approximation)

SSSSS [86.1K]

Answer:

9.412 rad/s.

Explanation:

Velocity is the rate of change of an object's position.

V = x/t

Where x is the distance in m

= 2.4 m

t is time taken in s

= 8.5 s

V = 2.4/8.5

= 0.2824 m/s.

Equating linear velocity and angular velocity,

V = ω*r

Where,

ω Is the angular speed in rad/s

r is the radius of the circle in m

= 3 cm

= 3cm * 1m/100 cm = 0.03 m

ω = V/r

= 0.2824/0.03

= 9.412 rad/s.

4 0

4 years ago

Logow se propong by

Lisa [10]

Answer:

I am very confused what your question is.

Explanation:

please clarify

7 0

3 years ago

A light source simultaneously emits light of two wavelengths, 480 nm and 560 nm, respectively. The source is used in a double-sl

Assoli18 [71]

Answer:

0.48 cm

Explanation:

given data

wavelength = 480 nm

wavelength = 560 nm

slit spacing = 0.040 mm

distance between double slits and the screen = 1.2 m

solution

we know that (1 nm= $10^{-9}$ m)

we wil take here equation of equations of interference that is

ym = R × (m λ)/d ..........................1

here m = 2 R i.e distance of screen and slit

so put here value and we get

separation between the second-order bright fringes = 0.48 cm

6 0

4 years ago