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3 years ago

Why is visible light safe to humans ?

Physics

2 answers:

MariettaO [177]3 years ago

7 0

Explanation:

Visible light is the range of electromagnetic waves that humans can see. Anything that glows or gives off light makes waves of what we call visible light. The sun, light bulbs, candles and fire are all things that people automatically think of when it comes to light, but visible light comes from many sources and in many colors. Some other sources of visible light include: television and computer screens, glow sticks, and fireworks

Neko [114]3 years ago

5 0

Answer:

Visible light is visible because most of the gases in planetary atmospheres (oxygen, nitrogen, carbon dioxide, methane, ammonia…) are transparent to those wavelengths. We evolved to see those wavelengths because the atmosphere in visible light is transparent.

Explanation:

You might be interested in

____________ are atoms with the same number of protons , but differing numbers of neutrons .

Masja [62]

ISOTOPES (C) are atoms with the same number of protons, but different numbers of neutrons.

:)

6 0

3 years ago

Calculate the work done by a 47 n force pushing a 0.025 kg pencil 0.25 m against a force of 23.

inna [77]

Fnet=fa-fr
=47-23
=24n
W=F∆Xcos°
=24n×0.025cos0
=0.6J

8 0

3 years ago

A satellite in the shape of a solid sphere of mass 1,900 kg and radius 4.6 m is spinning about an axis through its center of mas

algol [13]

Answer:

6.3 rev/s

Explanation:

The new rotation rate of the satellite can be found by conservation of the angular momentum (L):

$L_{i} = L_{f}$

$I_{i}*\omega_{i} = I_{f}*\omega_{f}$

The initial moment of inertia of the satellite (a solid sphere) is given by:

$I_{i} = \frac{2}{5}m_{s}r^{2}$

Where $m_{s}$ : is the satellite mass and r: is the satellite's radium

$I_{i} = \frac{2}{5}m_{s}r^{2} = \frac{2}{5}1900 kg*(4.6 m)^{2} = 1.61 \cdot 10^{4} kg*m^{2}$

Now, the final moment of inertia is given by the satellite and the antennas (rod):

$I_{f} = I_{i} + 2*I_{a} = 1.61 \cdot 10^{4} kg*m^{2} + 2*\frac{1}{3}m_{a}l^{2}$

Where $m_{a}$ : is the antenna's mass and l: is the lenght of the antenna

$I_{f} = 1.61 \cdot 10^{4} kg*m^{2} + 2*\frac{1}{3}150.0 kg*(6.6 m)^{2} = 2.05 \cdot 10^{4} kg*m^{2}$

So, the new rotation rate of the satellite is:

$I_{i}*\omega_{i} = I_{f}*\omega_{f}$

$\omega_{f} = \frac{I_{i}*\omega_{i}}{I_{f}} = \frac{1.61 \cdot 10^{4} kg*m^{2}*8.0 \frac{rev}{s}}{2.05 \cdot 10^{4} kg*m^{2}} = 6.3 rev/s$

Therefore, the new rotation rate of the satellite is 6.3 rev/s.

I hope it helps you!

5 0

4 years ago

Metal sphere 1 has a positive charge of 7.00 nc . metal sphere 2, which is twice the diameter of sphere 1, is initially uncharge

MariettaO [177]

Answer:

2.33 nC, 4.67 nC

Explanation:

when the two spheres are connected through the wire, the total charge (Q=7.00 nC) re-distribute to the two sphere in such a way that the two spheres are at same potential:

$V_1 = V_2$ (1)