Question

Alguem pode me ajudar nesses questoes?​

Answer 1

Resposta:

1- Fórmulas: °C/5 = °F-32/9 (Celcius p/ Fahrenheit e vice versa) °C = K-273 (Celcius p/ Kelvin e vice versa)

a)20/5=F-32/9 => 4=F-32/9 => 36 = °F-32 => °F=36+32 => °F=68

b)°C/5=9/9 => °C/5=1 => °C=5

c)°C=K-273

d)27=K-273 => K=273+27 => K= 300

e)°C=50-273 => °C=223

f) 41°F=5°C

5=K-273 => -K=-273-5 => K=273+5 => K= 278

g)°C=K-273 => °C=293-273 => °C=20

20°C=68°F

2-Resolve ai

3-Não sei

4-To=20°C / T=70 => ∆T=50°C

Lo=15m

Alfa=2,3.10-⁶

Fórmula => ∆L=Lo.Alfa.∆T => ∆L=15.2,3.10-⁶.50

∆L=1725.10-⁶ => ∆L=1,725.10³.10-⁶ => ∆L=1,725.10-³

∆L=0,001725m Novo comprimento=Lo+∆L

L=15+0,001725 => L=15,001725m

Cansei faz o resto vlw