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Evgesh-ka [11]
2 years ago
12

Two students walk in the same direction along a straight path, at a constant speed one at 0.90 m/s and the other at 1.90 m/s. a.

Assuming that they start at the same point and the same time, how much sooner does the faster student arrive at a destination 780 m away? b. How far would the students have to walk so that the faster student arrives 5.50 min before the slower student?
Physics
1 answer:
creativ13 [48]2 years ago
8 0

Answer: a) 456.66 s ; b) 564.3 m

Explanation: The time spend to cover any distance a constant velocity is given by:

v= distance/time so t=distance/v

The slower student time is: t=780m/0.9 m/s= 866.66 s

For the faster students t=780 m/1,9 m/s= 410.52 s

Therefore the time difference is 866.66-410.52= 456.14 s

In order to calculate the distance that faster student should  walk

to arrive 5,5 m before that slower student, we consider the follow expressions:

distance =vslower*time1

distance= vfaster*time 2

The time difference is 5.5 m that is equal to 330 s

replacing in the above expression we have

time 1= 627 s

time2 = 297 s

The distance traveled is 564,3 m

Guest
1 year ago
Plugging in 330 seconds from time 1 and time 2 into the distance equations would give you 627 and 297 meters not seconds.. unless I’m misunderstanding something?
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When two objects are in contact, moving together, which of the following statements must be true? Choose all that apply. When tw
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Answer:

The objects must have the same acceleration and the objects must exert the same magnitude force on each other.

Explanation:

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2 years ago
Driving your Ferrari through the Italian countryside at a speedy 88 m/s, you approach an opera diva singing a high C (1,046 Hz).
MrRissso [65]

Answer:

You will hear the note E₆

Explanation:

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The Doppler effect says that:

f' = \frac{v \pm v0 }{v \mp vs}*f

Where:

f = original frequency

f' = new frequency

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Replacing the values that we know in the equation we have:

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This frequency is close to the note E₆ (1,318.5 Hz)

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Answer:

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Explanation:

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