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Anon25 [30]
3 years ago
12

Alan leaves Los Angeles at 8:00 a.m. to drive to San Francisco, 400 mi away. He travels at a steady 45.0 mph . Beth leaves Los A

ngeles at 9:00 a.m. and drives a steady 55.0 mph .
a. Who gets to San Francisco first?
b. How long does the first to arrive have to wait for the second?
Physics
1 answer:
WARRIOR [948]3 years ago
6 0

Answer:

a.Beth

b.2232 s

Explanation:

We are given that

Distance,d=400 mi

Speed of Alan,v=45 mph

Speed of Beth,v'=55 mph

a.Time =\frac{distance}{speed}

Using the formula

Time taken by Alan=\frac{400}{45}=8.89 hr

Time taken by Beth=\frac{400}{55}=7.27hr

Alan will reach San Francisco at 4:53 PM

Beth will reach San Francisco at 4:16 PM

Beth will reach before Alan.

b.Difference between time=8.89-7.27=1.62 hr

t=1.62 hr

1.62-1=0.62 hr

0.62 hr=0.62\times 60\times 60=2232 s

Hence, Beth has to wait 2232 s for Alan to arrive .

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Carbon is allowed to diffuse through a steel plate 9.7-mm thick. The concentrations of carbon at the two faces are 0.664 and 0.3
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844°C

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<h3>What is Quadratic equation ?</h3>

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