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3 years ago

A student has a displacement of 304 m north in 180 s. What was the student's average velocity?

Physics

1 answer:

DanielleElmas [232]3 years ago

6 0

Answer:

v = 1.69 m/s

Explanation:

Given that,

Displacement of the student is 304 m due North and it takes 180 s.

We need to find the student's average velocity. Using formula of velocity.

Velocity = displacement/time

$v=\dfrac{304\ m}{180\ s}\\\\v=1.69\ m/s$

Hence, the student's average velocity is 1.69 m/s.

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3 years ago

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7nadin3 [17]

Answer:

22145.27733 ft

124984.76055 ft

Explanation:

The equation of pressure is

$P=P_0e^{-kh}$

where,

$P_0$ =Atmospheric pressure = 800 mbar

k = Constant

h = Altitude = 35000 ft

$P=\dfrac{1}{3}P_0$

$\dfrac{1}{3}P_0=P_0e^{-k35000}\\\Rightarrow \dfrac{1}{3}=e^{-k35000}\\\Rightarrow 3=e^{k35000}\\\Rightarrow ln3=k35000\\\Rightarrow k=\dfrac{ln3}{35000}\\\Rightarrow k=3.13\times 10^{-5}$

Now

$P=\dfrac{1}{2}P_0$

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The altitude will be 22145.27733 ft

$P=0.02P_0$

$0.02P_0=P_0e^{-kh}\\\Rightarrow 0.02=e^{-3.13\times 10^{-5}h}\\\Rightarrow ln0.02=-3.13\times 10^{-5}h\\\Rightarrow h=\dfrac{ln0.02}{-3.13\times 10^{-5}}\\\Rightarrow h=124984.76055\ ft$