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2 years ago

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A large fraction of the ultraviolet (UV) radiation coming from the sun is absorbed by the atmosphere. The main UV absorber in ou

r atmosphere is ozone, O3O3. In particular, ozone absorbs radiation with frequencies around 9.38×1014 HzHz . What is the wavelength λλlambda of the radiation absorbed by ozone? Express your answer in nanometers.
λ = ___.

2 answers:

vesna_86 [32]2 years ago

5 0

Answer:

0.03052 nm

Explanation:

From c = f¥

Where c = speed of electromagnetic wave in vacuum = 3x10^8 m/s

f = frequency of the wave

¥ = wavelenght of the wave.

¥ = c/f = (3x10^8)/(9.38×10^14)

= 3.052x10^-7 m

= 0.03052 nm

irakobra [83]2 years ago

3 0

Answer:

λ = 3.2 x 10⁻⁷ m = 320 nm

Explanation:

The relationship between the velocity of electromagnetic waves (UV rays) and the their frequency is:

v = fλ

where,

v = c = speed of the electromagnetic waves (UV rays) = speed of light

c = 3 x 10⁸ m/s

f = frequency of the electromagnetic waves (UV rays) = 9.38 x 10¹⁴ Hz

λ = wavelength of the electromagnetic waves (UV rays) = ?

Therefore, substituting the values in the relation, we get:

3 x 10⁸ m/s = (9.38 x 10¹⁴ Hz)(λ)

λ = (3 x 10⁸ m/s)/(9.38 x 10¹⁴ Hz)

<u>λ = 3.2 x 10⁻⁷ m = 320 nm</u>

So, the radiation of <u>320 nm</u> wavelength is absorbed by Ozone.

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The top floor of the Ostankino TV Tower in Moscow is located at a height of 360.4 m. Assume a stone is dropped from this top flo

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Answer:

$x=360.4-4.905t^2$

8.57181 s

84.0894561 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s² = a

Let distance from ground be x

From equation of motion we have

$s=ut+\frac{1}{2}at^2$

Here, distance covered while the stone is falling will be $360.4-x$

$360.4-x=ut+\frac{1}{2}at^2\\\Rightarrow 360.4-x=\frac{1}{2}9.81t^2\\\Rightarrow 360.4-x=4.905t^2\\\Rightarrow x=360.4-4.905t^2$

The equation is $x=360.4-4.905t^2$

At the ground x = 0

$0=360.4-4.905t^2\\\Rightarrow t=\sqrt{\dfrac{-360.4}{-4.905}}\\\Rightarrow t=8.57181\ s$

The time taken by the stone to fall to the ground is 8.57181 s

$v=u+at\\\Rightarrow v=0+9.81\times 8.57181\\\Rightarrow v=84.0894561\ m/s$

The velocity of the stone when it reaches the ground is 84.0894561 m/s

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2 years ago

The minute hand of a wall clock measures 16 cm from its tip to the axis about which it rotates. The magnitude and angle of the d

olya-2409 [2.1K]

Answer:

Explanation:

Given

Minute hand length =16 cm

Time at a quarter after the hour to half past i.e. 1 hr 45 min

Angle covered by minute hand in 1 hr is 360 and in 45 minutes 270

$|r|=\frac{3\times 2\pi r}{4}=75.408 cm$

$Angle =270^{\circ}$

(c)For the next half hour

Effectively it has covered 2 revolution and a quarter

$|r|=\frac{2\pi r}{4}=25.136 cm$

angle turned $=90^{\circ}$

(f)Hour after that

After an hour it again comes back to its original position thus displacement is same =25.136

Angle turned will also be same i.e. $90 ^{\circ}$

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Which of the following statements about horizons is true?

nalin [4]

<span>All soils have completely different horizon patterns.</span>

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1 year ago

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A 2.0 kg block on a horizontal frictionless surface is attached to a spring whose force constant is 590 N/m. The block is pulled

SVETLANKA909090 [29]

Answer:

The value is $v = -0.04 \ m/s$

Explanation:

From the question we are told that

The mass of the block is $m = 2.0 \ kg$

The force constant of the spring is $k = 590 \ N/m$

The amplitude is $A = + 0.080$

The time consider is $t = 0.10 \ s$

Generally the angular velocity of this block is mathematically represented as

$w = \sqrt{\frac{k}{m} }$

=> $w = \sqrt{\frac{590}{2} }$

=> $w = 17.18 \ rad/s$

Given that the block undergoes simple harmonic motion the velocity is mathematically represented as

$v = -A w sin (w* t )$

=> $v = -0.080 * 17.18 sin (17.18* 0.10 )$

=> $v = -0.04 \ m/s$

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2 years ago

A pyrotechnical releases a 3 kg firecracker from rest. at t=0.4 s, the firecracker is moving downward with a speed 4 m/s. At the

olga2289 [7]

Answer:

a) F = 30 N, b) I = 12 N s , c) I = -12 N s , d) ΔI = 0 N s

Explanation:

This exercise is a case at the moment, let's define the system formed by the firecracker and its two parts, in this case the forces during the explosion are internal and the moment is conserved

Initial, before the explosion

p₀ = m v

The speed can be found by kinematics

v = v₀ - g t

v = 0 - 10 0.4

v = -4.0 m / s

Final after division

pf = m₁ v₁f + m₂ v₂f

p₀ = pf

M v = m₁ v₁f + m₂ v₂f

Where M is the initial mass (M = 3 kg), m₁ is the mass mtop (m₁ = 1 kg) and m₂ in the mass m botton (m₂ = 2kg) and the piece that moves up (v₁f = 6m/s )

a) before the explosion the only force acting on the body is gravity

F = mg

F = 3 10 = 30 N

b) The expression for momentum is

I = Ft

Before the explosion the only force that acts is the weight

I = mg t

I = 3 10 0.4

I = 12 N s

c) To calculate this part we use the conservation of the moment and calculate the speed of the body that descends body 2

M v = m₁ v₁f + m₂ v₂f

v₂f = (M v - m₁ v₁f) / m₂

v₂f = (3 (-4) - 1 6) / 2

v₂f = - 9 m / 2

The negative sign indicates that body 2 (botton) is descending

Now we can use the momentum and momentum relationship for the body during the explosion

I = F t = Dp

F t = pf –po)

F t= [m₁ v₁f + m₂ v₂f]

I = [1 6 + 2 (-9) -0]

I = -12 N s

This is the impulse during the explosion the negative sign indicates that it is headed down

d) impulse change

I₀ = Mv

I₀ = 3 *4

I₀ =-12 N s

ΔI =If – I₀

ΔI = - 12 – (-12)

ΔI = -0 N s

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