Какое количество теплоты Q потребуется для обращения в воду льда массой 100 г, взятого при 0 °С и превращении всей воды в пар? Удельная теплота плавления льда λл = 3,4⋅105 Дж/кг, удельная теплота преобразования воды L = 2,3⋅106 Дж/кг. Удельная теплоёмкость воды c = 4200 Дж/(кг∙°С).

absorption lines are where light has been absorbed by the atom thus you see a dip in the spectrum whereas.

emission spectra have spikes in the spectra due to atoms releasing photons at those wavelengths.

explanation:

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.

i really dont know but i will try okay.

so i think u would just add all them up and get your answer..

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y