Please help asap that last part that’s blank

Substances which will glow in the dark after being exposed to sunlight are:

maw [93]

Luminescent I'm pretty sure :)

5 0

3 years ago

Measurement of heat conductivity A metal panel of area A = 100 cm² and thickness Y = 0.5 cm was found to conduct heat at a rate

RSB [31]

Answer:

The average thermal conductivity of the material at this temperature range is 3*10^(-9) W/mK.

Explanation:

The heat flow through this area of the metal panel can be expressed as:

$q/A=-k\frac{(T_2-T_1)}{Y}$

being A the surface area, k the thermal conductivity, T2 and T1 the temperatures in each side and Y the thickness of the metal panel.

We have to rearrange the equation to clear k:

$k=\frac{-(q/A)*Y}{\Delta T} = \frac{-(3 W/100 cm2)*0.5cm}{5 ^\circ K}*(\frac{1m}{100cm} )^{3} \\\\k=\frac{0.015}{5*100^{3} }W/mK= 3*10^{-9}W/mK$

The average thermal conductivity of the material at this temperature range is 3*10^(-9) W/mK.

7 0

3 years ago

Determine the energy of 1.70 mol of photons for each of the following kinds of light. (Assume three significant figures.)PART A

BabaBlast [244]

Answer:

For A: The energy of the given amount of photons for infrared radiation is $1.271\times 10^5J$

For B: The energy of the given amount of photons for infrared radiation is $4.026\times 10^5J$

For C: The energy of the given amount of photons for infrared radiation is $1.355\times 10^6J$

Explanation:

The relationship between energy and frequency is given by Planck's equation, which is:

$E=n\rimes N_A\times \frac{hc}{\lambda}$ ......(1)

where,

h = Planck's constant = $6.62\times 10^{-34}Js$

E = energy of the light

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength of light

$N_A$ = Avogadro's number = $6.022\times 10^{23}$

n = number of moles of photons = 1.70 moles

Conversion factor used: $1m=10^9nm$

For A:

Wavelength of infrared radiation = $1600nm=1.6\times 10^6m$

Putting values in equation 1, we get:

$E=1.7\times 6.022\times 10^{23}\times \frac{6.62\times 10^{-34}\times 3\times 10^8}{1.6\times 10^{-6}}\\\\E=1.271\times 10^5J$

Hence, the energy of the given amount of photons for infrared radiation is $1.271\times 10^5J$

For B:

Wavelength of visible light = $505nm=5.05\times 10^7m$

Putting values in equation 1, we get:

$E=1.7\times 6.022\times 10^{23}\times \frac{6.62\times 10^{-34}\times 3\times 10^8}{5.05\times 10^{-7}}\\\\E=4.026\times 10^5J$

Hence, the energy of the given amount of photons for infrared radiation is $4.026\times 10^5J$

For C:

Wavelength of ultraviolet radiation = $150nm=1.5\times 10^7m$

Putting values in equation 1, we get:

$E=1.7\times 6.022\times 10^{23}\times \frac{6.62\times 10^{-34}\times 3\times 10^8}{1.5\times 10^{-7}}\\\\E=1.355\times 10^6J$

Hence, the energy of the given amount of photons for infrared radiation is $1.355\times 10^6J$

3 0

4 years ago

The relative atomic mass of an element is the mass of its atoms compared with that of which element?

mariarad [96]

In calculating the relative atomic mass of an element with isotopes, the relative mass and proportion of each is taken into account. For example, naturally occurring chlorine consists of atoms of relative isotopic masses 35 (75%) and 37 (25%). Its relative atomic mass is 35.5.

5 0

4 years ago

Why is terrorism a law enforcement concern?

fomenos

Terrorism is a law enforcement concern because terrorism could cause harm to the communities they serve and terrorism can be a danger to the law enforcement it self. I hope I helped

4 0

3 years ago