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Anna71 [15]
3 years ago
5

Which of the following best represents potential energy being converted to kinetic energy? (2 points) The chemical energy in a b

attery is used to power a cell phone. The movement of wind turns a windmill that generates electricity. A man pedals a bicycle to ride across town. A woman kicks a soccer ball and scores a goal.
Chemistry
2 answers:
ziro4ka [17]3 years ago
6 0
A women kicks a soccer bal and scores a goal
valentina_108 [34]3 years ago
3 0

Answer: Option (d) is the correct answer.

Explanation:

Potential energy is defined as the energy present in an object due to its position.

Kinetic energy is the energy obtained by an object due to its motion. For example, a woman kicks a soccer ball and scores a goal.

When a work is done by an object due to its position or motion then this is known as mechanical energy. For example, the movement of wind turns a windmill that generates electricity.

When chemical energy is utilized to generate electrical energy in an electrochemical cell then it is known as electric current. For example, the chemical energy in a battery is used to power a cell phone.

Thus, we can conclude that a woman kicks a soccer ball and scores a goal, best represents potential energy being converted to kinetic energy.

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A thermometer having first-order dynamics with a time constant of 1 min is placed in a temperature bath at 100oF. After the ther
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(a) See below

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\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-0.5/1} \right )\\ & = & 100 + 10\left (1 - e^{-0.5} \right )\\ & = & 100 + 10 (1 - 0.6065)\\ & = & 100 + 10(0.3935)\\ & = & 100 + 3.935\\ & = & 103.935\,^{\circ}F\\\end{array}

(b) After 1 min

\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-1/1} \right )\\ & = & 100 + 10\left (1 - e^{-1} \right )\\ & = & 100 + 10 (1 - 0.3679)\\ & = & 100 + 10(0.6321)\\ & = & 100 + 6.321\\ & = & 106.321\,^{\circ}F\\\end{array}

2. Calculate the thermometer reading after 2.0 min

T₀ =106.321 °F

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  t = t - 1, because the cooling starts 1 min late

\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-(t - 1)/\tau} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-(2 - 1)/1} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-1} \right )\\ & = & 106.321 - 6.321 (1 - 0.3679)\\ & = & 106.321 - 6.321 (0.6321)\\ & = & 106.321 - 3.996\\ & = & 102.325\,^{\circ}F\\\end{array}

3. Plot the temperature readings as a function of time.

The graphs are shown below.

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