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4 years ago

Does visible light make up a relatively large part or a relatively small part of the electromagnetic spectrum?

Physics

1 answer:

Yakvenalex [24]4 years ago

8 0

Answer:

Relatively small part of the electromagnetic spectrum

Explanation:

Let us take a look at the electromagnetic spectrum

Radio waves

Wavelength = $10^3\ m$

Microwaves

Wavelength = $10^{-2}\ m$

Infrared waves

Wavelength = $10^{-5}\ m$

Visible waves

Wavelength = $0.5\times 10^{-6}\ m$

Radio waves

Wavelength = $10^3\ m$

Ultraviolet waves

Wavelength = $10^{-8}\ m$

X ray waves

Wavelength = $10^{-10}\ m$

Gamma ray waves

Wavelength = $10^{-12}\ m$

So, it can be seen that the visible light makes up a relatively small part of the electromagnetic spectrum.

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Why is it important for humans to help maintain the balance of the carbon cycle

gladu [14]

Because an excess of carbon dioxide could lead to global warming and affect climate drastically causing it to be warmer and destroy habitats such as penguins or polar bears.

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3 years ago

What is the equivalent resistance of the circuit?

Ivanshal [37]

Answer:

Since the wire is not splitting at any point in the circuit,

the resistors are in series

Hence, Equivalent resistance = 10 + 20 + 30

Equivalent Resistance = 60 Ω

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3 years ago

Alpha particles, each having a charge of +2e and a mass of 6.64 ×10-27 kg, are accelerated in a uniform 0.50 T magnetic field to

sergij07 [2.7K]

Answer:

$KE=1.2036\times 10^{-12}\ J$

Explanation:

Given:

charge on the alpha particle, $q=2e=3.2\times 10^{-19}\ C$

mass of the alpha particle, $m=6.64\times 10^{-27}\ kg$

strength of a uniform magnetic field, $B=0.5\ T$

radius of the final orbit, $r=0.5\ m$

<u>During the motion of a charge the magnetic force and the centripetal forces are balanced:</u>

$q.v.B=m.\frac{v^2}{r}$

$m.v=q.B.r$

where:

v = velocity of the alpha particle

$v=\frac{q.B.r}{m}$

$v=\frac{3.2\times 10^{-19}\times 0.5\times 0.5}{6.64\times 10^{-27}}$

$v=1.2048\times 10^{7}\ m.s^{-1}$

Here we observe that the velocity of the aprticle is close to the velocity of light. So the kinetic energy will be relativistic.

<u>We firstly find the relativistic mass as:</u>

$m'=\frac{1}{\sqrt{1-\frac{v^2}{c^2} } } \times m$

$m'=\frac{6.64\times 10^{-27}}{\sqrt{1-\frac{(1.2048\times 10^7)^2}{(3\times 10^8)^2} } }$

$m'=6.6533\times10^{-27}\ kg$

now kinetic energy:

$KE=m'.c-m.c$

$KE=6.6533\times 10^{-27}\times (3\times 10^8)^2-6.64\times 10^{-27}\times (3\times 10^8)^2$

$KE=1.2036\times 10^{-12}\ J$

6 0

3 years ago

a 42.3 kg girl and a 7.93 kg sled are on the surface of a frozen lake, 15.0m apart and linked by a rope, but not moving yet. the

ycow [4]

Answer:

they meet from the girl's original position at: 2.37 (meters)

Explanation:

We need to use the Newton's law, exactly the second law that relate force, mass and acceleration as: $F=m*a$ with this we can get both accelerations; solving for acceleration $a=\frac{F}{m}$ . Now $a_{girl}=\frac{5.76}{42.3}=0.14 (m/s^{2})$ and $a_{sled}=\frac{5.76}{7.93}=0.73(m/s^{2})$ . Then knowing that they both travel at the same time and assuming that the distance among the girl and the sled is: 15.0-x, so, $x=\frac{1}{2}*a_{girl}*t^{2}$ and $15.0-x=\frac{1}{2}*a_{sled}*t^{2}$ , solving for the time we get: $t=\sqrt{\frac{2x}{a_{girl} } }$ and $t=\sqrt{\frac{2*(15.0-x)}{a_{sled} } }$ with this equations we solving for the x that is the distance between the girl and the sled after the apply the force and we get: $\sqrt{\frac{2x}{a_{girl}}} = \sqrt{\frac{2*(15.0-x)}{a_{sled} }$ . Finally we get: $\frac{x}{a_{girl} }=\frac{(15.0-x)}{a_{sled} }$ and replacing the values we have got: $\frac{x}{0.14} =\frac{(15.0-x)}{0.73}$ so $5.33*x=15-x$ so x=2.37 (meters).