Answered

Westonci.ca is the trusted Q&A platform where you can get reliable answers from a community of knowledgeable contributors. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

this is a 3 part questionplease see picture for part A73) The human brain consumes about 22 W of power under normal conditions, though more power may be required during exams.(a) For what amount of time can one Snickers bar (see the note following Problem 48) power the normally functioning brain(one bar provides 280 calories)? (b) At what rate must you lift a 3.6-kg container of milk (one gallon) if the power output of your arm is to be 22 W? (C) How much time does it take to lift the milk container through a distance of 1.0 m at this rate?

This Is A 3 Part Questionplease See Picture For Part A73 The Human Brain Consumes About 22 W Of Power Under Normal Conditions Though More Power May Be Required class=

Sagot :

Given,

The power consumed by a brain, P=22 W

The energy per bar, E=280 calories=280Ɨ4184=1171.52 kJ

The mass of a milk container, m=3.6 kg

The power output of the arm, Pā‚€=22 W

The distance through which the container needs to be lifted, d=1.0 m

a)

The power is given by,

[tex]P=\frac{E}{t}[/tex]

Where t is the time.

On substituting the known values in the above equation,

[tex]\begin{gathered} 22=\frac{1171.52\times10^3}{t} \\ \Rightarrow t=\frac{1171.52\times10^3}{22} \\ =53250\text{ s} \end{gathered}[/tex]

That is,

[tex]\frac{53250}{3600}=14.79\text{ hr}[/tex]

Therefore one snicker bar can power the brain for 14.79 hr

b)

The power output can also be calculated using the formula,

[tex]\begin{gathered} P=F\times v \\ =mg\times v \end{gathered}[/tex]

Where F is the force applied by the container on the arm, v is the rate at which the container must be lifted, and g is the acceleration due to gravity.

On substituting the known values,

[tex]\begin{gathered} 22=3.6\times9.8\times v \\ v=\frac{22}{3.6\times9.8} \\ =0.62\text{ m/s} \end{gathered}[/tex]

Thus the rate at which the milk container must be lifted is 0.62 m/s

c)

The rate at which the container must be lifted is given by,

[tex]v=\frac{d}{t}[/tex]

Where t is the time it takes to lift the container at the calculated rate.

On substituting the known values,

[tex]\begin{gathered} 0.62=\frac{1}{t} \\ \Rightarrow t=\frac{1}{0.62} \\ =1.61\text{ s} \end{gathered}[/tex]

Thus it takes 1.61 s to lift the container through 1 m at the given rate.

Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.