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3 years ago

5

Ozone in the upper atmosphere absorbs energy in the 210-230 nmnm range of the spectrum. In what region of the electromagnetic sp

ectrum does this radiation occur

1 answer:

kifflom [539]3 years ago

5 0

Answer:

Ultraviolet

Explanation:

The electromagnetic spectrum is divided into a number of parts. One of which is the ultraviolet region. The ultraviolet region is found between the 100 nm - 400 nm band of the electromagnetic spectrum. From the range we just stated now, we can see for certainty that the given range 210 nm - 230 nm lies inside of the 100 nm c 400 nm of the electromagnetic spectrum, that of which is the ultraviolet region.

You might be interested in

The atom in the diagram has a neutral charge how many protons does it have

Olegator [25]

Explanation:

This should be of help:

An atom is made up of three subatomic particles:

Protons are the positively charged particles.
Electrons are the negatively charged particles.
Neutrons do not carry any charges.

A neutral atom is an atom that has not lost or gained any amount of electrons.

In a neutral atom;

the number of protons and electrons are the same

Usually atoms are designated this way:

ᵃₙX

where X is the symbol of the atom

a is the mass number of the atom

n is the atomic number of the atom

The mass number = number of protons + number of neutrons

Atomic number = number of protons

Note; atomic number is the same as the number protons and the number of electrons in a neutral atom.

Use this guide to solve the problem

Learn more:

Atomic number brainly.com/question/5425825

#learnwithBrainly

8 0

4 years ago

A0.350 kg iron horseshoe that is initially at 600°C is dropped into a bucket containing 21.9 kg of water at 21.8°C. What is the

Svetach [21]

Answer: $22.8^0C$

Explanation:-

$Q_{absorbed}=Q_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]$

where,

$m_1$ = mass of iron horseshoe = 0.35 kg = 350 g (1kg=1000g[/tex]

$m_2$ = mass of water = 21.9 kg = 21900 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of iron horseshoe = $600^oC$

$T_2$ = temperature of water = $21.8^oC$

$c_1$ = specific heat of iron horseshoe = $0.450J/g^0C$

$c_2$ = specific heat of water = $4.184J/g^0C$

Now put all the given values in equation (1), we get

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$

$350\times 0.450\times (T_{final}-600)^0C=-[21900g\times 4.184\times (T_{final}-21.8)]$

$T_{final}=22.8^0C$

Therefore, the final equilibrium temperature is $22.8^0C$ .

3 0

3 years ago

A 1.80-kg monkey wrench is pivoted 0.250 m from its center of mass and allowed to swing as a physical pendulum. The period for s

Evgen [1.6K]

Answer:

(a) I (Moment of inertia)=0.0987 $kgm^{2}$

(b) W(Angular Speed)=2.66 $\frac{rad}{s}$

Explanation:

Given data

m (Monkey mass)= 1.80 kg

d=2.50 m

T (Time Period)=0.940 s

Angle= 0.400 rad

(a) I (Moment of Inertia)=?

(b) W (Angular Speed)=?

For part (a) I (Moment of Inertia)=?

Time Period Formula is given as

$T=2(3.14)\sqrt{\frac{I}{mgd} }$

After Simplifying we get

$I=\frac{mgdT^{2} }{4*(3.14)^{2}}$

$I=\frac{1.8*9.8*0.25*(0.94)^{2} }{4(3.14)^{2} }$

$I=0.0987 kgm^{2}$

For Part (b) Angular Speed

From Kinetic Energy we get

$KE=\frac{1}{2}IW^{2}$

Pontential Energy

$PE=mgd(1-Cosa)$

KE=PE

$\frac{1}{2}IW^{2}=mgd(1-Cosa)$

$W^{2}=\frac{2mgd(1-Cosa)}{I}$

$W=\sqrt{\frac{2*1.8*9.8*0.25*(1-Cos(0.4)rad)}{0.0987} }$

$W=2.66\frac{rad}{s}$

5 0

3 years ago

A retaining wall is 5m long (into the plane of the paper), and it is supported by an anchor rod. The soil it supports has a weig

vekshin1

Knowing that the greatest horizontal pressure is the pressure made by the material on the wall -considering for that the total height of the wall-, then P = 4.375 t/m

---------------------

<u />

<u>Available data</u>:

The wall is 5m long into the plane ⇒ H

The soil density is 1.4 metric tonnes per cubic metre ⇒ γ

K = 0.25 ⇒ Coefficient of earth pressure. Depends on the angle.

From this information, we need to calculate the greatest horizontal pressure.

Horizontal pressure refers to the force made by the supported soil on the retaining wall that tends to deflect the wall outward.

We can calculate horizontal pressure at different heights from the top. And since we need to calculate the greatest horizontal pressure, we need to consider the total height that equals the wall height, H.

To do it, we will use the following formula, and replace the terms.

P = [ k γ H² ] / 2

P = [ 0.25 x 1.4 t/m³ x 5m² ] / 2

P = 8.75/2

P = 4.375 t/m

------------------------------------------

Related link: brainly.com/question/19381739?referrer=searchResults

brainly.com/question/12359444?referrer=searchResults

7 0

3 years ago

Amanda [17]

Answer:

The illumination on the book before the lamp is moved is 9 times the illumination after the lamp is moved.

Explanation:

The distance of the book before the lamp is moved, $d_{b} = 30 cm$

The distance of the book after the lamp is moved, $d_{a} = 90 cm$

Illumination can be given by the formula, $E = \frac{P}{4 \pi d^{2} }$

Illumination before the lamp is moved, $E_{b} = \frac{P}{4 \pi d_{b} ^{2} }$

Illumination after the lamp is moved, $E_{a} = \frac{P}{4 \pi d_{a} ^{2} }$

$\frac{E_{a}}{E_{b}} } = \frac{\frac{P}{4 \pi d_{a} ^{2} } }{\frac{P}{4 \pi d_{b} ^{2} } }$

$\frac{E_{a} }{E_{b} } = \frac{d_{b} ^{2} }{d_{a} ^{2}} \\\frac{E_{a} }{E_{b} } = \frac{30^{2} }{90 ^{2}}\\\frac{E_{a} }{E_{b} } =\frac{900 }{8100}\\\frac{E_{a} }{E_{b} } =\frac{1 }{9}$

$E_{b} = 9E_{a}$

The illumination on the book before the lamp is moved is 9 times the illumination after the lamp is moved.

6 0

3 years ago