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Sliva [168]
3 years ago
6

Why do people get sunburned by UV light ?

Physics
1 answer:
solong [7]3 years ago
8 0

Answer: Lack of melanin in the skin.

Explanation: The amount of melanin in your body is determined genetically. Some people dont produce enough melanin to protect their skin well. Eventually the UV causes the skin to burn, causing redness, swelling and pain.

PLEASE RATE 5 STARS AND VOTE AS BRAINLIEST:)

(^o^)(^o^)(^o^)(^o^)(^o^)(^o^)(^o^)(^o^)(^o^)(^o^)(^o^)(^o^)

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Imagine that a loudspeaker is producing a quiet tone with a low pitch. How will its vibrations change:
iVinArrow [24]
I believe it is A :) hope this helped
5 0
3 years ago
A bimetallic strip (brass/steel), which is straight at room temperature, will be immersed in boiling water and allowed to equili
lakkis [162]

The thermal expansion of the materials allows to find the deflection of the bimetallist strip is Δy = 3.48 cm

given paramers

    * Bimetallic brass / steel tape

    * Initial temperature, room temperature T = 20ºC

    * Final temperature, boiling water  = 100ºC

    * initial length L₀ = 222mm (1cm / 10mm) = 22.2cm

    * thickness of bimetallic tape e = 0.036 inch (2.54 cm/1 inch) = 0.0914 cm

to find

    * perpendicular deviation or deflection (Δy)

Thermal expansion is the phenomenon of change in the length of a body due to the change in temperature, due to the increase in the length of the atomic and molecular bonds, macroscopically it is described by

        ΔL = α L₀ ΔT

ΔL and ΔT are the variation of the length and temperature respectively, L₀ is the initial length and α the coefficient of expansion ends.

In this case we have a strip formed by two materials with different coefficient of thermal expansion,

Brass       α_{brass}   = 19 10⁻⁶ ºC⁻¹

Steel       α_{steel}    = 11 10⁻⁶ ºC⁻¹

In the attached we can see a diagram of the process, as the temperature increases, the material with greater thermal expansion lengthens more, so the system must curve towards the center of the material with less

thermal expansion. Let's find the length of the strip for each material

brass          L_{f brass} - L₀ = α_{brass} L₀ ΔT

Steel           L_{f steel} - L₀ = \alpha_{steel} L₀ ΔT

Note that the initial length is the same for the two materials and that the strip is in thermal equilibrium at room temperature.

If we assume that we have an arc of circumference, we can write the length of the arc

        θ = L / r

where θ is the angle in radines, L the length of the arc and r the radius of curvature, let's write this equation for each material

brass     L_{f \ brass} =θ r₁

steel      L_{f \ steel} = θ r₂

we substitute in our equations

           θ r₁ - L₀ = α_{brass} L₀ ΔT

           θ r₂ - L₀ = α_{steel} L₀ ΔT

let's subtract the two equations

           θ (r₁- r₂) = L₀ ΔT (α_{brass} - α_{steel})

the thickness of the strip is

           e = r₁ -r₂

           θ = Lo \ \Delta T \ \frac{\alpha_{brass} - \alpha_{steel}}{e}

we calculate

           θ = 22.2 \ (100-20) \ \frac{(19-11) \ 10^{-6}}{0.0914}

           θ = 0.155 rad

Let's use trigonometry to find the perpendicular deflection

          tan θ = Δy / L₀

          Δy = L₀ tan θ

          Δy = 22.2 tan 0.155

          Δy = 3.48 cm

Using the thematic expansion of the two materials we find the deflection of the bimetallist strip is 3.38 cm

Learn more about thermal expansion here: brainly.com/question/18717902

7 0
3 years ago
Which of these is an example of physical change?
My name is Ann [436]
That will be A melting a substance that's because the rest are examples of chemical changes and also its the same thing but in a different form.
8 0
3 years ago
Read 2 more answers
A weather reporter describes a large area of air that has the same temperature and air pressure . What is the reporter describin
Novay_Z [31]

<u>Answer:</u>

<u>air mass</u>

<u>Explanation:</u>

Remember, the weather of a particular location is often a description of the atmospheric condition of that particular place.

Indeed, the definition of an air mass fits the reporter's description above. As the name implies, <u>a particular air mass would represent the temperature characteristics of a particular area that has a constant or the same temperature and air pressure.</u>

5 0
3 years ago
A thin rod of length L and total charge Q has the nonuniform linear charge distribution λ(x)=λ0x/L, where x is measured from the
marissa [1.9K]

Answer:

Explanation:

λ(x) = λo x/ L

(a) The total charge is Q.

Q=\int_{0}^{L}dq

Q=\int_{0}^{L}\frac{\lambda _{0}x}{L}dx

Q=\frac{\lambda _{0}}{2L}\left ( x^{2} \right )_{0}^{L}

Q=\frac{\lambda _{0}}{2L}

λo = 2Q/L

(b)

Let at a distance x from the origin the charge is dq.

so, dq = (2Q/L) x/ L dx

dq=\frac{2Qx}{L^{2}}dx

The potential due to this small charge at a distance d to the left of origin

dV = \frac{KdQ}{d+x}

\int_{0}^{V}dV = \frac{2KQ}{L^{2}}\int_{0}^{L}\frac{xdx}{d+x}

V = \frac{2KQ}{L^{2}}\int_{0}^{L}\left ( 1- \frac{d}{d+x}\right )dx

V = \frac{2KQ}{L^{2}}\times \left ( x-dln(d+x) \right )\int_{0}^{L}

V = \frac{2KQ}{L^{2}}\times \left ( L-dln(d+L)-0+dlnd \right )

V = \frac{Q}{4\pi \epsilon _{0}L^{2}}\times \left ( L+d\times ln\left (\frac{d}{d+L}  \right )\right )

4 0
3 years ago
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