Question

6- Converta 380K para Celsius e Fahrenheit;

7- Converta -50 ºC para Fahrenheit e Kelvin;

8- Converta 160 ºF para Celsius e Kelvin.

Answer 1

6)

$\frac{380 - 273}{5} = \frac{tc}{5}$

Multiplica cruzado:

$tc.5 = 107.5$

$tc = 535 \div 5 = 107$

107°C

$\frac{380 - 273}{5} = \frac{tf - 32}{9}$

$(tf - 32).5 = 107.9$

$5tf - 160 = 963$

$5tf = 963 + 160$

$tf = 1123 \div 5$

$tf = 224.6$

224,6°F

7)

$\frac{ - 50}{5} = \frac{tf - 32}{9}$

$(tf - 32.5 )= - 50.9$

$5tf = - 450 + 160$

$tf = - 290 \div 5$

$tf = - 58$

-58°F

$\frac{ - 50}{5} = \frac{k - 273}{5}$

$(k - 273).5 = - 50.5$

$5k - 1365 = - 250$

$5k = - 250 + 1365$

$k = 1115 \div 5 = 223$

223k

8)

$\frac{160 - 32}{9} = \frac{tc}{5}$

$tc.9= 128 .5$

$tc = 640 \div 9$

$tc = 71.11$

71°C

$\frac{160 - 32}{9} = \frac{k - 273}{5}$

$(k - 273).9 = 128.5$

$9k - 2457 = 640$

$9k = 640 + 2457$

$k = 3097 \div 9$

$k = 344.11$

344k

Espero ter ajudado!

Answer 2

Explicação:

6- Converta 380K para Celsius e Fahrenheit;

$\dfrac{k - 273}{5} = \dfrac{c}{5}$

$\dfrac{380 - 273}{5} = \dfrac{c}{5}$

$\dfrac{107}{5} = \dfrac{c}{5}$

Eliminando denominadores iguais.

$107 = c$

c=>107⁰

________________________________

$\dfrac{k - 273}{5} = \dfrac{f - 32}{9}$

$\dfrac{380 - 273}{5} = \dfrac{f - 32}{9}$

$\dfrac{107}{5} = \dfrac{f - 32}{9}$

$21.4 = \dfrac{f - 32}{9}$

$f - 32 = 21.4 \times 9$

$f - 32 = 192.6$

$f = 192.6 + 32$

f=>224,6⁰

7- Converta -50 ºC para Fahrenheit e Kelvin;

$\dfrac{c}{5} = \dfrac{f - 32}{9}$

$- \dfrac{50}{5} = \dfrac{f - 32}{9}$

$- 10 = \dfrac{f - 32}{9}$

$f - 32 = - 10 \times 9$

$f - 32 = - 90$

$f = - 90 + 32$

f=>-58⁰

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$\dfrac{c}{5} = \dfrac{k - 273}{5}$

$- \dfrac{50}{5} = \dfrac{k - 273}{5}$

Eliminando denominadores iguais.

$- 50 = k - 273$

$- k = - 273 + 50$

$- k = - 223( - 1)$

k=>223⁰

8- Converta 160 ºF para Celsius e Kelvin.

$\dfrac{f - 32}{9} = \dfrac{c}{5}$

$\dfrac{160 - 32}{9} = \dfrac{c}{5}$

$\dfrac{128}{9} = \dfrac{c}{5}$

$9c = 128 \times 5$

$9c = 640$

$c = \dfrac{640}{9}$

c≈71,1⁰

________________________________

$\dfrac{f - 32}{9} = \dfrac{k - 273}{5}$

$\dfrac{160 - 32}{9} = \dfrac{k - 273}{5}$

$\dfrac{128}{9} = \dfrac{k - 273}{5}$

$9(k - 273) = 128 \times 5$

$9k - 2457 = 640$

$9k = 640 + 2457$

$9k = 3097$

$k = \dfrac{3097}{9}$

k=>344,1⁰