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rjkz [21]
3 years ago
15

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?
Physics
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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Answer:

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2 years ago
You drop an ice cube into an insulated bottle full of water and wait for the ice cube to completely melt. The ice cube initially
AVprozaik [17]

Answer:

T_{f}  = 17º C

Explanation:

This is a calorimetry problem, where heat is yielded by liquid water, this heat is used first to melt all ice, let's look for the necessary heat (Q1)  

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      Ice m = 80.0 g (1 kg / 1000 g) = 0.080 kg  

            L = 3.33 105 J / kg  

Water  M = 860 g = 0.860 kg  

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    Q₁ = m L  

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Now let's see what this liquid water temperature is when this heat is released  

      Q = M c_{e} ΔT = M c_{e} (T₀₁ -T_{f1})  

      Q₁ = Q  

     T_{f1} = T₀₁ - Q / M ce  

     T_{f1} = 26.0 - 2,664 10⁴ / (0.860 4186)  

     T_{f1} = 26.0 - 7.40  

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The initial temperature of water that has just melted is T₀₂ = 0ª  

The initial temperature of the liquid water is T₀₁= 18.6  

     m c_{e} T_{f} + M c_{e} T_{1} = M c_{e} T₀₁ - m c_{e} T₀₂o2  

         T_{f} = (M To1 - m To2) / (m + M)  

         T_{f} = (0.860 18.6 - 0.080 0) / (0.080 + 0.860)  

T_{f} = 17º C

 

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Explanation:

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