Question

Perform the following unit conversions:

a. 1 L to in.3
b. 650 J to Btu
c. 0.135 kW to ft · lbf/s
d. 378 g/s to lb/min
e. 304 kPa to lbf/in.2
f. 55 m3/h to ft3/s
g. 50 km/h to ft/s
h. 8896 N to ton (=2000 lbf)

Answer 1

Step-by-step explanation:

a. 1 liter (L) is equal to 61.0237 cubic inches (
`in^3`), So:

`1L*(61.0237in^3)/(1L)=61.0237in^3`

b. 1 J is equal to
`9.47817*10^(-4)` BTU. Thus:

`650J*(9.47817*10^(-4)BTU)/(1J)=6.16081*10^(-1)BTU`

c. 1 kW is equal to 737.56
`(ft\cdot lbf)/(s)`. So:

`0.135kW*(737.56(ft\cdot lbf)/(s))/(1kW)=99.57(ft\cdot lbf)/(s)`

d. 1
`(g)/(s)` is equal to 0.1323
`(lb)/(min)`. Therefore:

`378(g)/(s)*(0.1323(lb)/(min))/(1(g)/(s))=50.01(lb)/(min)`

e. 1 kPa is equal to 0.145
`(lbf)/(in^2)`. Thus:

`304kPa*(0.145(lbf)/(in^2))/(1kPa)=44.08(lbf)/(in^2)`

f. 1
`(m^3)/(h)` is equal to
`9.81*10^(-3)(ft^3)/(s)`. So:

`55(m^3)/(h)*(9.81*10^(-3)(ft^3)/(s))/(1(m^3)/(h))=5.40*10^(-1)(ft^3)/(s)`

g. 1
`(km)/(h)` is equal to 0.911
`(ft)/(s)`. Therefore:

`50(km)/(h)*(0.911(ft)/(s))/(1(km)/(h))=45.55(ft)/(s)`

h. 1 N is equal to
`1.1*10^(-4)` ton. Thus:

`8896N*(1.1*10^(-4)ton)/(1N)=0.98ton`