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3 years ago

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Infrared radiation has frequencies from 3.0×1011 to 3.0×1014 Hz, whereas the frequency region for radio wave radiation is 3.0×10

5 to 3.0×107 Hz. We can say that: 1. The speed of infrared radiation is radio wave radiation. 2. The wavelength of infrared radiation is radio wave radiation.

1 answer:

expeople1 [14]3 years ago

4 0

Answer:

1. The speed of infrared radiation is greater than radio wave radiation

2. The wavelength of infrared radiation is less than the radio wave radiation

Explanation:

Given:

The radio wave radiation N₁ is 3x10⁵ to 3x10⁷ Hz and the infrared radiation N₂ is 3x10¹¹ to 3x10¹⁴ Hz. The speed is:

$\frac{velocity_{2} }{velocity_{1} } =\frac{3x10^{11} }{3x10^{5} } to\frac{3x10^{14} }{3x10^{7} } =1x10^{6} to1x10^{7}$

1. The speed of infrared radiation is greater than radio wave radiation

2. The wavelength is:

λ = C/N

where more C, less λ

Then, the wavelength of infrared radiation is less than the radio wave radiation

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An electron of mass 9.11 1031 kg has an initial speed of 3.00 105 m/s. It travels in a straight line, and its speed increases to

elena55 [62]

Explanation:

It is given that,

Mass of an electron, $m=9.11\times 10^{-31}\ kg$

Initial speed of the electron, $u=3\times 10^5\ m/s$

Final speed of the electron, $v=7\times 10^5\ m/s$

Distance, d = 5 cm = 0.05 m

(a) The acceleration of the electron is calculated using the third equation of motion as :

$a=\dfrac{v^2-u^2}{2d}$

$a=\dfrac{(7\times 10^5)^2-(3\times 10^5)^2}{2\times 0.05}$

$a=4\times 10^{12}\ m/s^2$

Force exerted on the electron is given by :

$F=m\times a$

$F=9.11\times 10^{-31}\times 4\times 10^{12}$

$F=3.64\times 10^{-18}\ N$

(b) Let W is the weight of the electron. It can be calculated as :

$W=mg$

$W=9.11\times 10^{-31}\times 9.8$

$W=8.92\times 10^{-30}\ N$

Comparison,

$\dfrac{F}{W}=\dfrac{3.64\times 10^{-18}}{8.92\times 10^{-30}}$

$\dfrac{F}{W}=4.08\times 10^{11}$

Hence, this is the required solution.

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4 years ago

As his socks tumbled in the dryer, they became charged. If a small piece of lint with a charge of +1.25 E -19 C is attracted to

qwelly [4]

Answer:

$2.4\cdot 10^{10}N$

Explanation:

When an electric charge is immersed in an electric field, it experiences a force given by the equation

$F=qE$

where

q is the charge

E is the magnitude of the electric field

In this problem, we have:

$q=+1.25\cdot 10^{-19}C$ is the charge on the small piece of lint

$F=3.0\cdot 10^{-9}N$ is the force experienced by the charge

Therefore, we can find the magnitude of the electric field by re-arranging the equation and solving for E:

$E=\frac{F}{q}=\frac{3.0\cdot 10^{-9}}{1.25\cdot 10^{-19}}=2.4\cdot 10^{10}N$

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3 years ago

A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 N frictional force. He pushes in a dire

svp [43]

Answer:

(A). The work done on the cart by the friction is -700 J.

(B). The work done on the cart by the gravitational force is 0 J.

(C). The work done on the cart by the shopper is 700 J.

(D). The force the shopper exerts 38.61 N.

(E). The total work done on the cart is zero.

Explanation:

Given that,

Distance = 20.0 m

Frictional force = 35.0 N

Angle = 25.0°

(A). We need to calculate the work done on the cart by the friction

Using formula of work done

$W_{fr} = -F\cdot d$

Where, F = force

d = distance

Put the value into the formula

$W_{fr}=-35.0\times20$

$W_{fr}=−700\ J$

(B). The work done by the gravity is perpendicular to the direction of the motion

We need to calculate the work done on the cart by the gravitational force

Using formula of work done

$W=fd\cos\theta$

Put the value into the formula

$W=35.0\times20\cos90$

$W=0\ J$

(C). We need to calculate the work done on the cart by the shopper

Using formula of work done

$W_{sh}=W_{net}-W_{fr}$

Put the value into the formula

$W_{sh}=0-(-700)$

$W_{sh}=700\ J$

(D). We need to calculate the force the shopper exerts

Using formula of force

$F_{sh}=\dfrac{W_{fr}}{d\cos\theta}$

Put the value into the formula

$F_{sh}=\dfrac{700}{20\cos25}$

$F_{sh}=38.61\ N$

(E). We need to calculate the total work done on the cart

Using formula of work done

$W_{cart}=W_{fr}+W_{sh}$

Put the value into the formula

$W_{cart}=700-(-700)$

$W_{cart}=0\ J$

Hence, (A). The work done on the cart by the friction is -700 J.

(B). The work done on the cart by the gravitational force is 0 J.

(C). The work done on the cart by the shopper is 700 J.

(D). The force the shopper exerts 38.61 N.

(E). The total work done on the cart is zero.

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3 years ago

A rod 150 cm long and of diameter 2.0 cm is subjected to an axial pull of 20 kn. if the modulus of elasticity of the material of

Scilla [17]

Given:
rod of circular cross section is subjected to uniaxial tension.
Length, L=1500 mm
radius, r = 10 mm
E=2*10^5 N/mm^2
Force, F=20 kN = 20,000 N
[note: newton (unit) in abbreviation is written in upper case, as in N ]

From given above, area of cross section = π r^2 = 100 π =314 mm^2
(i) Stress,
σ
=force/area
= 20000 N / 314 mm^2
= 6366.2 N/mm^2
= 6370 N/mm^2 (to 3 significant figures)

(ii) Strain
ε
= ratio of extension / original length
= σ / E
= 6366.2 /(2*10^5)
= 0.03183
= 0.0318 (to three significant figures)

(iii) elongation
= ε * L
= 0.03183*1500 mm
= 47.746 mm
= 47.7 mm (to three significant figures)

5 0

4 years ago

An Airbus A320 jetliner has a takeoff mass of 75,000 kg. It reaches its takeoff speed of 82 m/s (180 mph) in 35 s. What is the t

timofeeve [1]

Answer:

175.5 kN

Explanation:

We find the acceleration of the Airbus A320 jetliner from,

a = (v - u)/t where u = initial velocity of jetliner = 0 m/s (since it starts from rest), v = final velocity of jetliner = 82 m/s and t = time for velocity change = 35 s

So a = (v - u)/t = (82 m/s - 0 m/s)/35 s = 82 m/s ÷ 35 s = 2.34 m/s²

Now, the thrust of the engines on the jetliner T = ma where m = mass of jetliner = 75,000 kg and a = acceleration of jetliner = 2.34 m/s²

T = ma

= 75,000 kg × 2.34 m/s²

= 175500 N

= 175.5 kN

8 0

3 years ago