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Neporo4naja [7]
3 years ago
14

If an aeroplane travels at 5000 m.p.h for an hour and 45 minutes, what distance does it travel?

Physics
1 answer:
adell [148]3 years ago
6 0

Answer:

8750 miles

Explanation:

The airplane travels with uniform motion (=constant velocity), so we can calculate the distance it travels using the formula for uniform motion:

d=vt

where

v is the speed of the plane

t is the time elapsed

In this problem:

v = 5000 mi/h is the speed of the plane

t = 1 h 45 min is the time elapsed

We have to convert this time into hours; we know that 45 minutes corresponds to 3/4 of an hour, that is 0.75 hours. So, the time here is

t=1h + 0.75 h = 1.75h

Therefore, the distance covered is:

d=(5000)(1.75)=8750 mi

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Answer:

V = 20.67 cm³

Explanation:

In this case, let's apply the Boyle's law which is:

P1V1 = P2V2

Where P1 and V1 would be condition in the water, and P2 and V2 would be the condition at the surface.

By logic, at the surface, pressure should be equals to 1 atm or 1.01x10^5 N/m²

We know the volume of the bubble at first which is 1.70 cm³ and we need to calculate V2. We know how much is P2, but we don't know the value of P1, which is the pressure of the bubble below the sea; this can be calculated using Pascal's principle which is the following expression:

P1 = Po + dgh

Where:

Po: innitial pressure, which we can assume is 1 atm

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Now replacing this data in the boyle's law we have the following:

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Replacing the data we have:

V2 = (1.01x10^5 + 1000*9.8*115) * 1.7 / 1.01x10^5

V2 = 2,087,600 / 1.01x10^5

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