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murzikaleks [220]
1 year ago
9

. (a) How long can you play tennis on the 800 kJ (about 200 kcal) of energy in a candy bar? (b) Does this seem like a long time?

Discuss why exercise is necessary but may not be sufficient to cause a person to lose weight. Answer from book:a) 30 or 32 minb) exercise does not burn calories, but requires a long time to burn off minor amounts of food.
Physics
1 answer:
galina1969 [7]1 year ago
8 0

(a). The power of the candy bar is,

P=440\text{ W}

The time taken to play on 800 kJ energy of the candy bar is,

t=\frac{E}{P}

where E is the energy.

Substituting the known values,

\begin{gathered} t=\frac{800\times10^3}{440} \\ t=1.818\times10^3\text{ s} \\ t=1818\text{ seconds} \end{gathered}

As 1 minute is equal to 60 seconds,

Thus,

\begin{gathered} t=1818\text{ seconds} \\ t=\frac{1818}{60} \\ t=30.3\text{ min} \end{gathered}

Thus, the time taken to play tennis on the 800 kJ energy is 30.3 minutes.

(b). By doing the exercise, the process of digestion of food inside our body increases. Thus, the exercise does not helps us to burn the calories. But it helps us to diggest the heavy meal like candy bar easily.

The time taken to digest the canndy bar or to utilise its energy is large because it takes a lot of time to burn small amount of food and make it digest quickly.

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A reconnaissance plane flies 560 km away from its base at 602 m/s, then flies back to its base at 903 m/s.
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Approximately 722\; \rm m\cdot s^{-1}.

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The average speed of a vehicle is calculated as:

\displaystyle \text{average speed} = \frac{\text{total distance}}{\text{total time}}.

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The unit of the speeds in this question is meters per second, while the unit of distance is kilometers. Convert the unit of distance to meters:

560 \; \rm km = 560 \times 10^{3} \; \rm m = 5.6 \times 10^{5}\; \rm m.

1120 \; \rm km = 1120 \times 10^{3} \; \rm m = 1.12 \times 10^{6}\; \rm m.

Time required for the first part of this trip:

\displaystyle \frac{5.60 \times 10^{5}\; \rm m}{602\; \rm m\cdot s^{-1}} \approx 930\; \rm s.

Time required for the second part of this trip:

\displaystyle \frac{5.60 \times 10^{5}\; \rm m}{903\; \rm m\cdot s^{-1}} \approx 620\; \rm s.

The time required for the entire trip would be approximately 930 + 620 = 1550\; \rm s.

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