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salantis [7]
3 years ago
6

7. A golfer hits a golf ball a through a horizontal displacement of 80m. The time of flight for the ball was 1.8

Physics
1 answer:
Darya [45]3 years ago
6 0

Answer:

44.4 m/s

Explanation:

d = 80 m, t = 1.8 s, find v

d = v*t

v = d/t = 80/1.8 = 44.4 m/s

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The air in a room with volume 180 m3 contains 0.25% carbon dioxide initially. fresher air with only 0.05% carbon dioxide flows i
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Answer:

p(t)=\frac{1}{20}(1+2e^{-t/90})

Explanation:

Let p(t) be % of CO_2 at time t(time in minutes).

Amount of CO_2 in room=Volume of room \times p

Rate of change of CO_2 in room=Volume of room x \frac{dp}{dt}

=180\times \frac{dp}{dt}

Rate of inflow of CO_2=(\% CO_2 \ in\  Fresh\  Air \times \frac{1}{100})\times(Rate \ of  Inflow\ of \ air)

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Rate of CO_2 outflow

(\% CO_2 \ in\  Fresh\  Air \times \frac{1}{100})\times(Rate \ of  Outflow\ of \ air)\\=(p\times\frac{1}{100})\times2=\frac{p}{50}

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% rate of change is100\times \ Rate of \ Change=\frac{1}{10}-2p

Therefore we have:

180\frac{dp}{dt}=\frac{1}{10}-2p\\\frac{dp}{dt}=\frac{1-20p}{1800}\\\\\frac{dp}{20p-1}=-\frac{dt}{1800}

Integrate both:

\int\limits \frac{dp}{20p-1}=\int\limits-\frac{dt}{1800}

\frac{1}{20}In|20p-1|=-\frac{t}{1800}+In \ C

In|20p-1|=-\frac{t}{90}+InK

+20In \ C=+In \ K

Raise both sides to base e,

20p-1=e^{-\frac{t}{90}+In \ K}\\\\p=\frac{1}{20}(1+Ke^{-t/90})

Initially, there's only 0.25% of CO_2,

Now substitute p=0.25 and t=0

0.25=\frac{1}{20}(1+K)\\K=4

p=\frac{1}{20}(1+2e^{-t/90})

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