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Vinvika [58]
3 years ago
12

The box of a well-known breakfast cereal states that one ounce of the cereal contains 113 Calories (1 food Calorie = 4186 J). If

1.79% of this energy could be converted by a weight lifter's body into work done in lifting a barbell, what is the weight of the heaviest barbell that could be lifted a distance of 1.90 m?
Physics
2 answers:
zimovet [89]3 years ago
5 0

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);

work done=mgh..................(1)

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;

E=mgh.....................(2)

Also, it should be recalled that the weight, W of an object is given by the equation below;

W=mg...............(3)

Substituting (3) into (2) we obtain the following;

E=Wh..................(4)

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;

E=\frac{1.79}{100}*473018J=8467.02J

Putting this value of E into equation (4), we obtain the following;

8467.02=W*1.9\\

therefore;

W=\frac{8467.02}{1.9}=4456.33N

This is the heaviest weight of barbell the lifter could lift with the amount of energy specified.

mafiozo [28]3 years ago
3 0

Answer:

m=454.73 kg

Explanation:

To convert the energy of the cereal to know the weight:

m*g*h=Q'

Cal=0.0179*113 Cal=2.0227Cal

Q'=2.0227Cal*\frac{4186J}{1Cal}

m*g*h=8467.02J

m*g=\frac{Q'}{h}=\frac{8467.02J}{1.90m}

m*g=4456.33 N

Supuse the gravity as a g=9.8m/s^2

m=\frac{4456.33N}{9.8m/s^2}

m=454.73 kg

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Answer: 247.67 V

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Given

Potential At A V_a=382\ V

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when particle starts from A it reaches with velocity v_b at Point while when it starts from C it reaches at point B with velocity 2v_b

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