Ask question

Ask question

All categories

2 years ago

12

Two of the types of ultraviolet light, uva and uvb, are both components of sunlight. their wavelengths range from 320 to 400 nm

for uva and from 290 to 320 nm for uvb. compare the energy of microwaves, uva, and uvb. rank from greatest to least energy per photon. to rank items as equivalent, overlap them.

2 answers:

a_sh-v [17]2 years ago

7 0

In light energy, the higher the frequency, the greater the energy a light contain.

We know for a certain that frequency is just the reciprocal of wavelength:

frequency = 1 / wavelength

Calculating for frequencies:

f UVA = 1/320 to 1/400

f UVA = 0.0031 to 0.0025

f UVB = 1/290 to 1/320

f UVB = 0.0034 to 0.0031

Since UVB has higher frequency range, then it has higher energy than UVA.

Rufina [12.5K]2 years ago

3 0

Answer:

uvb > uva > microwaves

Explanation:

The energy of an electromagnetic wave is

$E =\frac{hc}{\lambda}$

Where

h is the Planck's constant

c is the speed of light and

$\lambda$ is the frequency

So we can understand that the energy is inversely proportional to the wavelength

The wavelengths are

1. microwaves - 1 mm - 1m

2. uva - 320 - 400 nm

3. uvb - 290 - 320 nm

So with the smallest wavelength the uvb has the greatest energy

and the microwave has least amount of energy with the largest wavelength range

so the answer is uvb > uva > microwaves

You might be interested in

Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work i

Elis [28]

Four electrons are placed at the corner of a square

So we will first find the electrostatic potential at the center of the square

So here it is given as

$V = 4\frac{kQ}{r}$

here

r = distance of corner of the square from it center

$r = \frac{a}{\sqrt2}$

$r = \frac{10nm}{\sqrt2} = 7.07 nm$

$Q = e = -1.6 * 10^{-19} C$

now the net potential is given as

$V = \frac{4 * 9*10^9 * (-1.6 * 10^{-19})}{7.07 * 10^{-9}}$

$V = 0.815 V$

now potential energy of alpha particle at this position

$U_i = qV = 2*1.6 * 10^{-19} * (-0.815) = -2.6 * 10^{-19} J$

Now at the mid point of one of the side

Electrostatic potential is given as

$V = 2\frac{kQ}{r_1} + 2\frac{kQ}{r_2}$

here we know that

$r_1 = \frac{a}{2} = 5 nm$

$r_2 = \sqrt{(a/2)^2 + a^2} = \frac{\sqrt5 a}{2}$

$r_2 = 11.2 nm$

now potential is given as

$V = 2\frac{9 * 10^9 * (-1.6 * 10^{-19})}{5 * 10^{-9}} + 2\frac{9*10^9 * (-1.6 * 10^{-19})}{11.2 * 10^{-9}}$

$V = -0.576 - 0.257 = -0.833 V$

now final potential energy is given as

$U_f = q*V = 2*1.6 * 10^{-19}* (-0.833) = -2.67 * 10^{-19} J$

Now work done in this process is given as

$W = U_f - U_i$

$W = (-0.267 * 10^{-19}) - (-0.26 * 10^{-19}}$

$W = -7 * 10^{-22} J$

8 0

2 years ago

At what time t is the turtle second time a distance of 10.0 cm from its starting point?

skad [1K]

Answer:

10 cm.

Explanation:

5 0

2 years ago

What is the radius of an automobile tire that turns with a period of 0.091 s and has a linear

faust18 [17]

Answer:

1) The radius of the tire is approximately 0.28966 meters

2) The centripetal force is the force that keeps a body moving on a circular path

Explanation:

1) The linear speed of the automobile tire = 20.0 m/s

The period with which the tire turns = 0.091 s

The period = The time it takes to make a complete turn

Therefore;

The number of turns in 1 second = 1/0.091 ≈ 10.989 turns

The distance covered with 10.989 turns, assuming no friction = 10.989 × The circumference of the tire

∴ The distance covered with 10.989 turns, assuming no friction = 10.989 × 2 × π × Radius of the tire

From the speed of the car, 20.0 m/s, we have;

The distance covered in 1 second = 20.0 meters

Therefore;

10.989 × 2 × π × Radius of the tire = 20.0 meters

Radius of the tire = (20.0 meters)/(10.289 × 2 × π) ≈ 0.28966 meters

The radius of the tire ≈ 0.28966 meters

2) The centripetal force is the force required to maintain the curved motion of an object, and having a direction towards the center of the rotary motion.

The centripetal force is given by the formula, $F = \dfrac{m \cdot v^2}{r}$

Where;

F = The centripetal force

m = The mass of the object

v = The linear velocity of the object

r = The radius of the rotational motion.

3 0

2 years ago

A 60-kg motor sits on four cylindrical rubber blocks. Each cylinder has a height of 3 cm and a cross-sectional area of 15 cm2. T

stealth61 [152]

<span>A.) If a sideways force of 300 N is applied to the motor, how far will it move sideways?</span>

3 0

2 years ago

Read 2 more answers

Two convex thin lenses with focal lengths 10.0 cm and 20.0 cm are aligned on a common axis, running left to right, the 10-cm len

love history [14]

Answer:

(c) +6.67

Explanation:

f1 = 10 cm

f2 = 20 cm

u = Object distance = 15 cm

Distance between lenses = 20 cm

For first lens image distance

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{10}-\frac{1}{15}\\\Rightarrow \frac{1}{v}=\frac{1}{30}\\\Rightarrow v=30\ cm$

Distance from second lens is 10 cm to the right

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{20}-\frac{1}{-10}\\\Rightarrow \frac{1}{v}=\frac{3}{20}\\\Rightarrow v=6.67\ cm$

The final image will appear as +6.67 cm

3 0

2 years ago