PHYSICS COLLEGE

Ultraviolet light is typically divided into three categories. UV-A, with wavelengths between 400 nm and 320 nm, has been linked with malignant melanomas. UV-B radiation, which is the primary cause of sunburn and other skin cancers, has wavelengths between 320 nm and 280 nm. Finally, the region known as UV-C extends to wavelengths of 100 nm. (a) Find the range of frequencies for UV-B radiation. (b) In which of these three categories does radiation with a frequency of 7.9 * 1014 Hz belong

Answers

Answer 1

Answer:

Answer:

a) The UV-B has frequencies between $9.375x10^(14)Hz$ and $1.071x10^(15)Hz$

b) The radiation with a frequency of $7.9x10^(14)Hz$ belong to the UV-A category.

Explanation:

(a) Find the range of frequencies for UV-B radiation.

Ultraviolet light belongs to the electromagnetic spectrum, which distributes radiation along it in order of different frequencies or wavelengths.

Higher frequencies:

Gamma ray

X ray

Ultraviolet rays

Visible region

Lower frequencies:

Infrared

Microwave

Radio waves

That radiation is formed by electromagnetic waves, which are transverse waves formed by an electric field and a magnetic field perpendicular to it. Any of those radiations will have a speed of $3x10^{8]m/s$ in vacuum.

The velocity of a wave can be determined by means of the following equation:

$c = \nu \cdot \lambda$ (1)

Where c is the speed of light, $\nu$ is the frequency and $\lambda$ is the wavelength.

Then, from equation 1 the frequency can be isolated.

$\nu = (c)/(\lambda)$ (2)

Before using equation 2 to determine the range of UV-B it is necessary to express $\lambda$ in units of meters in order to match with the units from c.

$\lambda = 320nm . (1m)/(1x10^(9)nm)$ ⇒ $3.2x10^(-7)m$

$\lambda = 280nm . (1m)/(1x10^(9)nm)$ ⇒ $2.8x10^(-7)m$

$\nu = (3x10^(8)m/s)/(3.2x10^(-7)m)$

$\nu = 9.375x10^(14)s^(-1)$

$\nu = 9.375x10^(14)Hz$

$\nu = (3x10^(8)m/s)/(2.8x10^(-7)m)$

$\nu = 1.071x10^(15)Hz$

Hence, the UV-B has frequencies between $9.375x10^(14)Hz$ and $1.071x10^(15)Hz$

(b) In which of these three categories does radiation with a frequency of $7.9x10^(14)Hz$ belong.

The same approach followed in part A will be used to answer part B.

Case for UV-A:

$\lambda = 400nm . (1m)/(1x10^(9)nm)$ ⇒ $4x10^(-7)m$

$\nu = (3x10^(8)m/s)/(4x10^(-7)m)$

$\nu = 7.5x10^(14)s^(-1)$

$\nu = 7.5x10^(14)Hz$

Hence, the UV-A has frequencies between $7.5x10^(14)Hz$ and $9.375x10^(14)Hz$ .

Therefore, the radiation with a frequency of $7.9x10^(14)Hz$ belongs to UV-A category.

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An object of mass m swings in a horizontal circle on a string of length L that tilts downward at angle θ.Find an expression for the angular velocity ω in terms of g, L and angle θ.

Answers

Answer:

The expression would be ω = $\sqrt{(g)/(L sin 0 ) }$

Explanation:

Given that ω is the angular velocity

g is the acceleration due to gravity

L is the length

θ is the angle of downward tilt

For an object we compare the horizontal and vertical component of the forces acting on the body;

For vertical component

T sinθ = mg............1

For the horizontal component

T cos θ = $(mv^(2) )/(R)$ .............2

R is our radius and is = L cos θ

v = ωR

substituting into equation 2 we have

T cos θ = m(ωR $)^(2)$ /R

T cos θ=m(ω $)^(2)$ R ..................3

Now comparing the vertical and the horizontal component we have;

equation 1 divided by equation 3 we have

T sin θ /T cos θ = mg / m(ω $)^(2)$ R

Tan θ = g / (ω $)^(2)$ R............4

Making ω the subject formula we have;

(ω $)^(2)$ = g/ R Tan θ

But R = L cos θ and Tan θ = sin θ/ cosθ

putting into equation 4 we have;

(ω $)^(2)$ = g /[( L cos θ) x( sin θ/ cosθ)]

(ω $)^(2)$ = g/ L sinθ

ω = $\sqrt{(g)/(L sin 0 ) }$

Therefor the expression for the angular velocity ω in terms of g, L and angle θ would be ω = $\sqrt{(g)/(L sin 0 ) }$

Let g, r, L and T are gravity, radius, length, and angle of string w/r/t vertical, respectively.
Then
ω²r = rg/Lcos T
ω² = g/L cos T
ω = √(g / L cos T)

An emf is induced by rotating a 1207 turn, 20.0 cm diameter coil in the Earth's 4.13 10-5 T magnetic field. What average emf is induced, given the plane of the coil is originally perpendicular to the Earth's field and is rotated to be parallel to the field in 10.0 ms

Answers

Answer:

0.157 V

Explanation:

Parameters given:

Number of turns, N = 1207

Diameter of coil = 20 cm = 0.2 m

Radius of coil, r = 0.2/2 = 0.1 m

Magnetic field strength, B = $4.13 * 10^(-5) T$

Time interval, t = 10 ms = $10 * 10^(-3) = 0.01 s$

The average EMF induced in a coil due to a magnetic field is given as:

EMF = $(- N * A * B)/(t)$

where A = Area of coil

A = π $r^(2)$

Therefore, EMF will be:

$EMF = (- 1207 * 3.142 * 0.1^2 * 4.13 * 10^(-5))/(0.01) \n\n\nEMF = -0.157 V\n$

Chapter 38, Problem 001 Monochromatic light (that is, light of a single wavelength) is to be absorbed by a sheet of a certain material. Photon absorption will occur if the photon energy equals or exceeds 0.42 eV, the smallest amount of energy needed to dissociate a molecule of the material.

(a) What is the greatest wavelength of light that can be absorbed by the material?
(b) In what region of the electromagnetic spectrum is this wavelength located?

Answers

Answer:

a) $\lambda=2.95x10^(-6)m$

b) infrared region

Explanation:

Photon energy is the "energy carried by a single photon. This amount of energy is directly proportional to the photon's electromagnetic frequency and is inversely proportional to the wavelength. If we have higher the photon's frequency then we have higher its energy. Equivalently, with longer the photon's wavelength, we have lower energy".

Part a

Is provide that the smallest amount of energy that is needed to dissociate a molecule of a material on this case 0.42eV. We know that the energy of the photon is equal to:

$E=hf$

Where h is the Planck's Constant. By the other hand the know that $c=f\lambda$ and if we solve for f we have:

$f=(c)/(\lambda)$

If we replace the last equation into the E formula we got:

$E=h(c)/(\lambda)$

And if we solve for $\lambda$ we got:

$\lambda =(hc)/(E)$

Using the value of the constant $h=4.136x10^(-15) eVs$ we have this:

$\lambda=(4.136x10^(15)eVs (3x10^8 (m)/(s)))/(0.42eV)=2.95x10^(-6)m$

$\lambda=2.95x10^(-6)m$

Part b

If we see the figure attached, with the red arrow, the value for the wavelenght obtained from part a) is on the infrared region, since is in the order of $10^(-6)m$

Why might a scientist want to use a model to study the solar system? O A. Its extreme simplicity makes it difficult to see patterns in observations. B. Its extremely slow movement makes it difficult to see the motions of different planets. C. Its extremely large size makes it difficult to see all of its parts at the same time. D. Its extremely small size makes it difficult to see planets that are far away

Answers

A scientist wants to use a model to study the solar model because its extremely large size makes it difficult to see all of its parts at the same time. Hence, option C is correct.

What is a Solar System?

The Sun and all the smaller movable objects that orbit it make up the Solar System. The eight main planets are the largest objects in the Solar System, excluding the Sun. Mercury, Venus, Earth, and Mars are the four relatively tiny, rocky planets closest to the Sun.

The asteroid belt, which is home to millions of stony objects, lies beyond Mars. These are remains from the planets' creation 4.5 billion years ago.

Jupiter, Saturn, Uranus, and Neptune are the four gas giants that can be found on the opposite side of the asteroid belt. Despite being much larger than Earth, these planets are rather light. Their main components are hydrogen and helium.

To get more information about Solar systems:

brainly.com/question/18365761

#SPJ5

If my weight on Earth is 140lbs, what is my mass?

Answers

Answer:

63.57 kg

Explanation:

weight = 140 lbs

Let the mass is m.

1 lbs = 4.45 N

The weight of an object is defined as the force with which our earth attracts the body towards its centre.

Weight is the product of mass of the body and the acceleration due to gravity of that planet.

W = m x g

On earth surface g = 9.8 m/s^2

Now convert lbs in newton

So, 140 lbs = 140 x 4.45 = 623 N

So, m x 9.8 = 623

m = 63.57 kg

Thus, the mass is 63.57 kg.

Two violin strings are tuned to the same frequency 294 H. The tension in one string is then decreased by 2.0%. What will be the beat frequency heard when the two strings are played together?