Question

GENTE POR FAVOR ME AJUDEM É PARA AMANHÃ E TEM QUE FAZER A FORMULA\ CONTA MAS EU NÃO SEI ME AJUDEM POR FAVOR

converta para a escala fahrenheit a) 35 C \ B) 50 C \ C) -58 C \ D)250K \ E)368 K \ F)313 K \\\ converta para escala kelvin a)27 C \ b) -230C\ c) 273 C \ d)50F \ e)122 F \ f)82,4 F \ converta para escala celsius 25 K \ 395 K

Answer 1

converta para a escala Fahrenheit

A

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

$\frac{35}{5} = \frac{tf - 32}{9}$

5tf - 160 = 315

5tf = 315 + 160

5tf = 475

tf = $\frac{475}{5}$

tf = 95 °F

B

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

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

5tf - 160 = 450

5tf = 450 + 160

5tf = 610

tf = $\frac{610}{5}$

tf = 122 °F

C

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

$\frac{- 58}{5} = \frac{tf - 32}{9}$

5tf - 160 = -522

5tf = -522 + 160

5tf = -362

tf = $\frac{-362}{5}$

tf = -72,4 °F

D

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

$\frac{250 - 273}{5} = \frac{tf - 32}{9}$

$\frac{-23}{5} = \frac{tf - 32}{9}$

5tf - 160 = -207

5tf = -207 + 160

5tf = -47

tf = $\frac{-47}{5}$

tf = -9,4 °F

E

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

$\frac{368 - 273}{5} = \frac{tf - 32}{9}$

$\frac{95}{5} = \frac{tf - 32}{9}$

5tf - 160 = 855

5tf = 855 + 160

5tf = 1015

tf = $\frac{1015}{5}$

tf = 203 °F

F

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

$\frac{313- 273}{5} = \frac{tf - 32}{9}$

$\frac{40}{5} = \frac{tf - 32}{9}$

5tf - 160 = 360

5tf = 360 + 160

5tf = 520

tf = $\frac{520}{5}$

tf = 104 °F

converta para escala Kelvin

A

K = tc + 273

K = 27 + 273

K = 300

B

K = tc + 273

K = -230 + 273

K = 43

C

K = tc + 273

K = 273 + 273

K = 546

D

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

$\frac{K - 273}{5} = \frac{50 - 32}{9}$

$\frac{K - 273}{5} = \frac{18}{9}$

9K - 2457 = 90

9K = 90 + 2457

9K = 2547

K = $\frac{2547}{9}$

K = 283

E

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

$\frac{K - 273}{5} = \frac{122 - 32}{9}$

$\frac{K - 273}{5} = \frac{90}{9}$

9K - 2457 = 450

9K = 450 + 2457

9K = 2907

K = $\frac{2907}{9}$

K = 323

F

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

$\frac{K - 273}{5} = \frac{82,4 - 32}{9}$

$\frac{K - 273}{5} = \frac{50,4}{9}$

9K - 2457 = 252

9K = 252 + 2457

9K = 2709

K = $\frac{2709}{9}$

K = 301

converta para escala Celsius

K = tc + 273

25 = tc + 273

-tc = 273 - 25

-tc = 248 · (-1)

tc = -248

K = tc + 273

395 = tc + 273

-tc = 273 - 395

-tc = -122 · (-1)

tc = 122