contestada

Which of the following would be a correct use of dimensional analysis to convert 45 mph to feet per sec? a. [tex] \frac{45 \: miles}{hour} \times \frac{1mile}{5280ft} \times \frac{1hour}{60min} \times \frac{1min}{60sec} [/tex]b. [tex] \frac{45miles}{hour} \times \frac{5280ft}{1mile} \times \frac{1hour}{60min} \times \frac{1min}{60sec} [/tex]c. [tex] \frac{45miles}{1hour} \times \frac{1ft}{5280miles} \times \frac{1hour}{60min} \times \frac{1min}{60sec} [/tex]

Respuesta :

We have the following conversion regarding that:

1 mile = 5280.02 feet

1 hour = 3600 sec

Then, to convert 45 mph to feet per sec, we have:

[tex]45\frac{m}{h}\cdot5280.02\frac{f}{m}\cdot\frac{1h}{3600s}\Rightarrow\frac{45\cdot5280.02}{3600}ft/\sec [/tex]

The units m/m = 1, h/h =1, and then the units remaining were feet / sec.

Then, we have:

[tex]\frac{45\cdot5280.02}{3600}=66.00025ft/\sec [/tex]

Then, the two conversions are:

• 45 m/h * 5280.0 feet/mile, and

,

• 1 h/ 3600sec = (1 hour/60min) * (1 min/60sec).

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