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3 years ago

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Explain how you can use Boyle's law to determine the new volume of gas when its pressure is increased from 270kPa to 540kPa? The

orginal volume of gas was 1L. Assume the temperature and number of particles are constant. What is the new volume?

1 answer:

ivann1987 [24]3 years ago

4 0

I hope this is what you are looking for P1 = 270 kPa V1 = 1 L P2 = 540 kPa V2 = ? Use Boyle's Law: P1V1 = P2V2 V2 = P1 V1 / P2 V2 = 270 x 1 / 540 V2 = 0.5 L

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A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

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Answer:

The rate of change of the area when the bottom of the ladder (denoted by $b$ ) is at 36 ft. from the wall is the following:

$\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s$

Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

$A=\sqrt{l^2-b^2}b$

The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

$\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}$

$\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Product rule

$\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Chain rule

$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

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3 years ago

Which action best demonstrates the transformation of mechanical energy to heat energy and sound energy?

MrMuchimi

Answer:

striking a hammer on a nail

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3 years ago

A gas has an initial volume of 212 cm^3 at a temperature of 293 K and a pressure of 0.98 atm. What is the final pressure of the

MatroZZZ [7]

For this we use general equation for gases. Our variables represent:

p- pressure
v-volume
t- temperature

P1V1/T1 = P2V2/T2

in this equation we know:
P1,V1 and T1, T2 and V2.
We have one equation and 1 unknown variable.

P2 = T2P1V1/T1V2 = 1.1atm

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3 years ago

Motor oil , with a viscosity of 0 . 250 Ns / m2 , is flowing through a tube that has a radius of 5 . 00 mm and is 25 . 0 cm long

Thepotemich [5.8K]

Answer:

1.1775 x 10^-3 m^3 /s

Explanation:

viscosity, η = 0.250 Ns/m^2

radius, r = 5 mm = 5 x 10^-3 m

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According to the Poisuellie's formula

Volume flow per unit time is

$V=\frac{\pi \times Pr^{4}}{8\eta l}$

$V=\frac{3.14 \times 300000\times \left ( 5\times 10^{-3} \right )^{4}}{8\times 0.250\times 0.25}$

V = 1.1775 x 10^-3 m^3 /s

Thus, the volume of oil flowing per second is 1.1775 x 10^-3 m^3 /s.

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3 years ago

8. How did the measured angular magnification of the telescope compare with the theoretical prediction?

Genrish500 [490]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The theoretical angular magnification lies within the angular magnification range

Explanation:

From the question we are told that

The focal length of B is $f_{objective } = 43.0 \ cm$

The focal length of A is $f_{eye} = 10.4 \ cm$

The theoretical angular magnification is mathematically represented as

$m = \frac{f_{objective }}{f_{eye}} = \frac{43.0}{10.4}$

$m = \frac{f_{objective }}{f_{eye}} = 4.175$

Form the question the measured angular magnification ranges from 4 -5

So from the value calculated and the value given we can deduce that the theoretical angular magnification lies within the angular magnification range

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3 years ago