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2 years ago

Why do waves with frequencies higher than visible light hurt while those with lower frequencies do not affect us?

2 answers:

7 0

Lower frequency waves have less strength to penetrate. How bad a wave is all depends on how well it penetrates our bodies. Visible light doesn't penetrate ur skin, but UV rays (higher than visible) can go through our skin, making it <span>bad" for us. High frequency waves have more energy and move faster</span>

AnnZ [28]2 years ago

4 0

Answer:

Explanation:

Waves with higher frequency have high penetration power. Energy of a wave id directly proportional to the frequency of the wave so penetration power is high. Visible rays suits to the human body and do not harm. Humans can see visible light and can easily work in visible range. But in the higher frequency such as UV waves and radio waves penetrate human body and affect it in negative way.

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The burning of a log releases the logs chemical_energy into other forms of energy

saul85 [17]

Answer:

When we burn wood we are releasing solar energy, in the form of heat, that has been stored in the wood as chemical energy. The process of photosynthesis converted solar energy, water and carbon dioxide into oxygen and the organic molecules that form the wood, half the weight of which is carbon.

Explanation:

7 0

3 years ago

Describe what happens to the particles of water when it boils.

Nezavi [6.7K]

Answer:

The liquid turns to a gas.

Explanation:

If a liquid is heated the particles are given more energy and move faster and faster expanding the liquid. Particles in the middle of the liquid form bubbles of gas in the liquid.

8 0

3 years ago

A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s

aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is $U_{s}$ = 0

And, initial potential gravitational potential energy of the rod is $U_{g} = \frac{mgL}{2}$ .

It is given that,

mass of the bar = 0.795 kg

g = 9.8 $m/s^{2}$

L = length of the rod = 0.2 m

Initial total energy T = $\frac{mgL}{2}$

Now, when the rod is in horizontal position then final total energy will be as follows.

T = $\frac{1}{2}kx^{2} + I \omega^{2}$

where, I = moment of inertia of the rod about the end = $\frac{mL^{2}}{3}$

Also, $\omega = \frac{\nu}{L}$

where, $\nu$ = speed of the tip of the rod

x = spring extension

The initial unstrained length is $x_{o}$ = 0.1 m

Therefore, final length will be calculated as follows.

x' = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m

Then, x = $x' - x_{o}$

x = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m - 0.1 m

= 0.1236 m

k = 25 N/m

So, according to the law of conservation of energy

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}$

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

Putting the given values into the above formula as follows.

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

$\frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}$

v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

7 0

3 years ago

Calculate the energy of the green light emitted, per photon, by a mercury lamp with a frequency of 5.49 × 1014 hz.

Tcecarenko [31]

The energy of a photon is given by
$E=hf$
where
$h=6.6 \cdot 10^{-34} Js$ is the Planck constant
f is the frequency of the photon

In our problem, the frequency of the light is
$f=5.49 \cdot 10^{14}Hz$
therefore we can use the previous equation to calculate the energy of each photon of the green light emitted by the lamp:
$E=hf=(6.6 \cdot 10^{-34}Js)(5.49 \cdot 10^{14} Hz)=3.62 \cdot 10^{-19} J$

8 0

3 years ago

when two object P and Q are supplied with the same quantity of heat, the temperature change in P is observed to be twice that of

Ahat [919]

<h2>When two object P and Q are supplied with the same quantity of heat, the temperature change in P is observed to be twice that of Q. The mass of P is half that of Q. The ratio of the specific heat capacity of P to Q</h2>

Explanation:

Specific heat capacity

It is defined as amount of heat required to raise the temperature of a substance by one degree celsius .

It is given as :

Heat absorbed = mass of substance x specific heat capacity x rise in temperature

or ,

Q= m x c x t

In above question , it is given :

For Q

mass of Q = m

Temperature changed =T₂/2

Heat supplied = x

Q= mc t

X=m x C₁ X T₁

or, X =m x C₁ x T₂/2

or, C₁=X x 2 /m x T₂ (equation 1 )

For another quantity : P

mass of P =m/2

Temperature= T₂

Heat supplied is same that is : X

so, X= m/2 x C₂ x T₂

or, C₂=2X/m. T₂ (equation 2 )

Now taking ratio of C₂ to c₁, We have

C₂/C₁= 2X /m.T₂ /2X /m.T₂

so, C₂/C₁= 1/1

so, the ratio is 1: 1

8 0

3 years ago