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siniylev [52]
2 years ago
13

A single Oreo cookie provides 53 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without accelerat

ion) a 100-kilogram weight 2-decimeters above the ground with an energy efficiency of 25%. How many repetitions can she do with the energy supplied from a single Oreo cookie? What happens to the number of repetitions that can be done if the efficiency increases?
Physics
1 answer:
Sever21 [200]2 years ago
4 0

Answer:

Approximately 325 (rounded down,) assuming that g = 9.81\; {\rm N \cdot kg^{-1}}.

The number of repetitions would increase if efficiency increases.

Explanation:

Ensure that all quantities involved are in standard units:

Energy from the cookie (should be in joules, {\rm J}):

\begin{aligned} & 53\; {\rm kCal} \times \frac{1\; {\rm kJ}}{4.184\; {\rm kCal}} \times \frac{1000\; {\rm J}}{1\; {\rm kJ}} \approx 2.551 \times 10^{5}\; {\rm J} \end{aligned}.

Height of the weight (should be in meters, {\rm m}):

\begin{aligned} h &= 2\; {\rm dm} \times \frac{1\; {\rm m}}{10\; {\rm dm}} = 0.2\; {\rm m}\end{aligned}.

Energy required to lift the weight by \Delta h = 0.2\; {\rm m} without acceleration:

\begin{aligned} W &= m\, g\, \Delta h \\ &= 100\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 0.2\; {\rm m} \\ &= 196\; {\rm N \cdot m} \\ &= 196\; {\rm J} \end{aligned}.

At an efficiency of 0.25, the actual amount of energy required to raise this weight to that height would be:

\begin{aligned} \text{Energy Input} &= \frac{\text{Useful Work Output}}{\text{Efficiency}} \\ &= \frac{196\; {\rm J}}{0.25} \\ &=784\; {\rm J}\end{aligned}.

Divide 2.551 \times 10^{5}\; {\rm J} by 784\; {\rm J} to find the number of times this weight could be lifted up within that energy budget:

\begin{aligned} \frac{2.551 \times 10^{5}\; {\rm J}}{784\; {\rm J}} &\approx 325 \end{aligned}.

Increasing the efficiency (the denominator) would reduce the amount of energy input required to achieve the same amount of useful work. Thus, the same energy budget would allow this weight to be lifted up for more times.

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3 years ago
Suppose two point charges, Q1 = -6.25 x 10-9 C and Q2 = -6.25 x 10-9 C, are separated by a distance d = 0.617 m.
n200080 [17]

Answer:

The answer to your question is the letter A) F =  9.23 x 10⁻⁷ N

Explanation:

Data

q₁ = -6.25 x 10⁻⁹ C

q₂ = -6.25 x 10⁻⁹ C

d = 0.617 m

k = 9 x 10⁹ Nm²/C²

F = ?

Formula

              F = k q₁q₂ /r²

-Substitution

              F = (9 x 10⁹)(-6.25 x 10⁻⁹)(-6.25 x 10⁻⁹) / (0.617)²

-Simplification

              F = 3.512 x 10⁻⁷ / 0.381

-Result

              F = 9.227 x 10⁻⁷ N  ≈ 9.23 x 10⁻⁷ N

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An increase in resistance causes the flow of electric current to
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An increase in voltage causes the flow of electric current to INCREASE. An increase in resistance causes the flow of electric current to DECREASE.

100% right

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what is the difference between kinetic energy and random motion of a cylider of oxygen carried by a car and one standing on a pl
pochemuha

Answer:

See explanation below

Explanation:

If we are talking about the kinetic energy of the cylinder of oxygen:

The kinetic energy possessed by any object is given by

K=\frac{1}{2}mv^2

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In this case, we have one cylinder carried by a car and one standing on a platform: this means that the speed of the cylinder carried by the car will be different from zero (and so also its kinetic energy will be different from zer), while the speed of the cylinder standing on the platform will be zero (and so its kinetic energy also zero). Therefore, the kinetic energy of the cylinder carried by the car will be larger than that standing on a platform.

Instead, if we are talking about the kinetic energy due to the random motion of the molecules of oxygen inside the cylinder:

The kinetic energy of the molecules in a gas is directly proportional to the absolute temperature of the gas:

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Interactive Solution 9.37 presents a method for modeling this problem. Multiple-Concept Example 10 offers useful background for
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