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jolli1 [7]
3 years ago
8

The nutritional calorie (Calorie) is equivalent to 1 kcal. One pound of body fat is equivalent to about 4.1 x 103 Calories. If a

runner expends 1850 kJ/h, how long (h) does she have to run to work off 7.56 lb of body fat? Enter your answer in decimal format with one decimal place.
Physics
1 answer:
Luba_88 [7]3 years ago
5 0

Answer:

t = 70.13 hours

Explanation:

As we know that 1 lb fat can be used to expend 4.1 k Calorie of energy

so here we know that we have to use 7.56 lb of body fat

so here the total energy that is to be utilized is given as

Q = (4.1 \times 10^3)(7.56)

so we will have

Q = 30996 Calorie

now we know that runner expends the energy at rate of 1850 kJ/h

so we have

E = 1850 \times 10^3 kJ/h

so it is

E = 0.44 \times 10^3 calorie/h

now we know that time to burn this energy is given as

t = \frac{Q}{E}

t = \frac{30996}{0.44 \times 10^3}

t = 70.13 hours

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Answer:

1. 75N

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Explanation:

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3 years ago
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2 years ago
Which of the following statements are true of thermal energy and kinetic energy?
Irina-Kira [14]
This question is asking you to determine if individual atoms or systems, or both have these types of energy. A system would be "all the molecules or atoms" whereas an individual atom is "each of the molecules or atoms."

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7 0
2 years ago
Suppose you first walk 12.0 m in a direction 20? west of north and then 20.0 m in a direction 40.0? south of west. how far are y
Gnesinka [82]
The representation of this problem is shown in Figure 1. So our goal is to find the vector \overrightarrow{R}. From the figure we know that:

\left | \overrightarrow{A} \right |=12m \\ \\ \left | \overrightarrow{B} \right |=20m \\ \\ \theta_{A}=20^{\circ} \\ \\ \theta_{B}=40^{\circ}

From geometry, we know that:

\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}

Then using vector decomposition into components:

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So if you want to find out <span>how far are you from your starting point you need to know the magnitude of the vector \overrightarrow{R}, that is:
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</span>\theta_R=180^{\circ}+tan^{-1}(\frac{\left | R_y \right |}{\left | R_x \right |}) \\ \\ \theta_R=180^{\circ}+tan^{-1}(\frac{1.58}{19.42}) \\ \\ \theta_R=180^{\circ}+4.65^{\circ}=185.85^{\circ}


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