Respuesta :
Answer:
[tex]m=454.73 kg[/tex]
Explanation:
To convert the energy of the cereal to know the weight:
[tex]m*g*h=Q'[/tex]
[tex]Cal=0.0179*113 Cal=2.0227Cal[/tex]
[tex]Q'=2.0227Cal*\frac{4186J}{1Cal}[/tex]
[tex]m*g*h=8467.02J[/tex]
[tex]m*g=\frac{Q'}{h}=\frac{8467.02J}{1.90m}[/tex]
[tex]m*g=4456.33 N[/tex]
Supuse the gravity as a g=9.8m/s^2
[tex]m=\frac{4456.33N}{9.8m/s^2}[/tex]
[tex]m=454.73 kg[/tex]
Answer:
4456.33N
Explanation:
Let the energy converted by the lifter's body into work done be E.
The work done in lifting the barbell is against gravity and it is given by equation (1);
[tex]work done=mgh..................(1)[/tex]
where m is the mass of the barbell, g is acceleration due to gravity and h is the vertical distance (height) through which it is lifted. Hence by the principle of energy conversion as specified by the problem we could write the following equation;
[tex]E=mgh.....................(2)[/tex]
Also, it should be recalled that the weight, W of an object is given by the equation below;
[tex]W=mg...............(3)[/tex]
Substituting (3) into (2) we obtain the following;
[tex]E=Wh..................(4)[/tex]
Given; h = 1.9m
To obtain the energy E converted by the lifter's body in Joules, we use the specifications given in the problem as follows;
Given; 1cal = 4186J
therefore 113cal = (113 x 4186)J = 473018J. This is the total energy contained in one ounce of cereal consumed by the lifter. However as specified by the problem, only 1.79% of this energy was converted by the lifter's body into weight lifting, therefore;
[tex]E=\frac{1.79}{100}*473018J=8467.02J[/tex]
Putting this value of E into equation (4), we obtain the following;
[tex]8467.02=W*1.9\\[/tex]
therefore;
[tex]W=\frac{8467.02}{1.9}=4456.33N[/tex]
This is the heaviest weight of barbell the lifter could lift with the amount of energy specified.