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3 years ago

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One ounce of a well-known breakfast cereal contains 103 Calories (1 food Calorie = 4186 J). If 1.6% of this energy could be conv

erted by a weight lifter's body into work done in lifting a barbell, what is the heaviest barbell that could be lifted a distance of 1.8 m?

1 answer:

faust18 [17]3 years ago

5 0

Answer:

the heaviest barbell that could be lifted is 390.6kg

Explanation:

Hello!

To solve this question you must follow the following steps

1. Find the amount of energy that can transform the body into motion, this is achieved by multiplying the breakfast consumed by the percentage of energy conversion.

$W=(103Calories)(0.016)\frac{4186J}{1calorie} =6898.5J$

2. We use the equation that defines the work done by a body that has weight when it is lifted, this is defined by the product of mass by gravity by height.

W=mgh

where

W=work=6898.5J

m=mass

g=gravity=9.81m/s^2

h=height=1.8m

now we solve for mass, and use the values.

$m=\frac{W}{hg} =\frac{6898.5}{(1.8)(9.81)} =390.6kg$

the heaviest barbell that could be lifted is 390.6kg

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Explanation:

a.

Speed is a scalar quantity while velocity is a vector quantity.

Speed is defined as rate of change of distance per unit time whereas velocity is defined as rate of change of displacement per unit time.

Speed is a directionless quantity while velocity constitutes direction.

b.

<em>Total time of round trip when we're given:</em>

distance travelled to the right, $d_r=90000\ m$

speed while travelling to the right, $v_r=25\ m.s^{-1}$

time spent at gas station, $t_g=600\ s$

time spent while travelling back towards the left, $t_l=30\times 60=1800\ s$

speed while travelling to the left, $v_{_l}=20\ m.s^{-1}$

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$t_r=\frac{d_r}{v_r}$

$t_r=\frac{90000}{25}$

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<u>Therefore total time taken in the round trip:</u>

$t=t_r+t_l+t_g$

$t=3600+600+1800$

$t=6000\ s$

c.

<em>Now, distance travelled towards left:</em>

$d_l=v_{_l}\times t_l$

$d_l=20\times1800$

$d_l=36000\ m$

<u>Therefore total distance:</u>

$d=d_l+d_r$

$d=36000+90000$

$d=126000\ m$

d.

Now, total displacement:

$s=d_r-d_l$

$s=90000-36000$

$s=54000\ m$ towards right from the starting point.

e.

<u>Average speed:</u>

$v_a=\frac{d}{t}$

$v_a=\frac{126000}{6000}$

$v_a=21\ m.s^{-1}$

f.

<u>Average velocity:</u>

$\vec v_a=\frac{s}{t}$

$\vec v_a=\frac{54000}{6000}$

$\vec v_a=9\ m.s^{-1}$ towards right.

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