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statuscvo [17]
3 years ago
11

A sample of gas has an initial volume of 23.6 l at a pressure of 1.52 atm . if the sample is compressed to a volume of 10.5 l: ,

what will its pressure be? (assume constant temperature.)
Physics
1 answer:
nataly862011 [7]3 years ago
7 0

New pressure=3.42 atm

Explanation:

using Boyle's law

P1 V1= P2 V2

P1= initial pressure

V1= initial volume

P2= final pressure

V2=final volume

23.6 (1.52)=P2 (10.5)

P2=3.42 atm

so final pressure= 3.42 atm

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8 0
3 years ago
Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.20×10^6 N, one at an angle 14.0∘ west of north,
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3 years ago
An above ground swimming pool of 30 ft diameter and 5 ft depth is to be filled from a garden hose (smooth interior) of length 10
STALIN [3.7K]

This question involves the concepts of dynamic pressure, volume flow rate, and flow speed.

It will take "5.1 hours" to fill the pool.

First, we will use the formula for the dynamic pressure to find out the flow speed of water:

P=\frac{1}{2}\rho v^2\\\\v=\sqrt{\frac{2P}{\rho}}

where,

v = flow speed = ?

P = Dynamic Pressure = 55 psi(\frac{6894.76\ Pa}{1\ psi}) = 379212 Pa

\rho = density of water = 1000 kg/m³

Therefore,

v=\sqrt{\frac{2(379212\ Pa)}{1000\ kg/m^3}}

v = 27.54 m/s

Now, we will use the formula for volume flow rate of water coming from the hose to find out the time taken by the pool to be filled:

\frac{V}{t} = Av\\\\t =\frac{V}{Av}

where,

t = time to fill the pool = ?

A = Area of the mouth of hose = \frac{\pi (0.015875\ m)^2}{4} = 1.98 x 10⁻⁴ m²

V = Volume of the pool = (Area of pool)(depth of pool) = A(1.524 m)

V = [\frac{\pi (9.144\ m)^2}{4}][1.524\ m] = 100.1 m³

Therefore,

t = \frac{(100.1\ m^3)}{(1.98\ x\ 10^{-4}\ m^2)(27.54\ m/s)}\\\\

<u>t = 18353.5 s = 305.9 min = 5.1 hours</u>

Learn more about dynamic pressure here:

brainly.com/question/13155610?referrer=searchResults

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Answer:

it depends on a person's own weight

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2 years ago
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