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alexdok [17]
3 years ago
6

Which type of waves from the electromagnetic spectrum are used in heat lamps and heat sensing devices?

Physics
2 answers:
Alex73 [517]3 years ago
7 0
Infrared waves are used in heat lamps and other heat sensing devices. Infrared waves or commonly known as Infrared radiations (IR) is the type of electromagnetic radiation we encounter most in our everyday life. Heat lamps are electrical devices which emit infrared radiation.
Anuta_ua [19.1K]3 years ago
6 0

Answer:

Infrared radiations (IR) is what use daily in our lives.

Explanation:

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If it goes 1250km every hour, times 4.2 hours
1250*4.2
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Find the westward component of a resultant vector 85.42 unit, 23 degrees W of N
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Since the angle is West of North, therefore to find for the westward component (horizontal component) of the vector, we use the sin function:

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So the westward component (x) is:

x = 85.42 sin 23

<span>x = 33.38 unit</span>

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 A student needs help completing an energy transfer diagram for a ceiling fan. Which BEST fills in the blanks labeled M and N to
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On August 10, 1972, a large meteorite skipped across the atmosphere above the western United States and western Canada, much lik
Anon25 [30]

a) 4.62\cdot 10^{14} J

b) 0.110 megatons

c) 8.46 bombs

Explanation:

a)

The energy lost by the meteorite is equal to the difference between its final kinetic energy and its initial kinetic energy:

\Delta K=K_f-K_i

Which can be rewritten as:

\Delta K=\frac{1}{2}mv^2-\frac{1}{2}mu^2

where:

m=3.2\cdot 10^6 kg is the mass of the meteorite

v=0 is the final speed of the meteorite

u=17 km/s = 17,000 m/s is the initial speed of the meteorite

Substituting the values into the equation, we found the loss in energy of the meteorite:

\Delta K=0-\frac{1}{2}(3.2\cdot 10^6)(17000)^2=-4.62\cdot 10^{14} J

So, the energy lost by the meteorite is 4.62\cdot 10^{14} J

b)

The energy equivalent to 1 megaton of TNT is

E_{TNT}=4.2\cdot 10^{15} J

Here the energy lost by the meteorite is

E=4.62\cdot 10^{14} J

Therefore, in order to write the energy lost by the meteorite as a multiple of the energy of 1 megaton of TNT, we have to divide the energy lost by the meteorite by the energy equivalent to 1 TNT; we find:

\frac{E}{E_{TNT}}=\frac{4.62\cdot 10^{14}}{4.2\cdot 10^{15}}=0.110

So, the energy lost by the meteorite corresponds to 0.110 megatons.

c)

The energy of one atomic bomb explosion in Hiroshima is equal to

E'=13 kt (13 kilotons)

which corresponds to

E'=0.013 Mt (0.013 megatons)

Here the energy of the meteorite is equal to

E=0.110 Mt (0.110 megatons)

Therefore, we can find how many Hiroshima bombs are equivalent to teh meteorite impact by using the following rules of three:

\frac{1 bomb}{0.013 Mt}=\frac{x bombs}{0.110 Mt}\\x=\frac{1\cdot 0.110}{0.013}=8.46

So, 8.46 bombs.

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C the atmosphere provides warmth
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