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3 years ago

14

The pressure of a gas is 15 atm in a 5L cylinder. If the volume of the cylinder is depressed to 3L, What is the new pressure exe

rted by the gas?

1 answer:

disa [49]3 years ago

5 0

P1v1=p2v2. 15x5=p2x3

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Explanation:

In addition to stand-alone discussion boards, all of these sites include discussion boards as part of their features except:

A. Linkedln

All other application in addition to stand-alone Google, Yahoo, Microsoft, etc. include discussion board.

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Un movil aumenta su velocidad de 10m/s a 20m/s acelerando uniformemente a razon de 5m/s2 ¿que distancia logro en dicha operacion

Kruka [31]

v₀ = initial velocity of the mobile = 10 m/s

v = final velocity of the mobile = 20 m/s

a = acceleration of the mobile = 5 m/s²

d = distance traveled during this operation = ?

Using the kinematics equation

v² = v²₀ + 2 a d

inserting the above values in the equation

20² = 10² + 2 (5) d

400 = 100 + 10 d

subtracting 100 both side

400 - 100 = 100 - 100 + 10 d

300 = 10 d

dividing both side by 10

300/10 = 10 d/10

d = 30 m

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3 years ago

An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.50 mm. A 25

torisob [31]

Answer:

The electric field is 16666.66 V/m.

The surface charge density is $1.474\times10^{-7}\ C/m^2$

The capacitance is $4.484\times10^{-12}\ F$

The charge on each plate is $112.1\times10^{-12}\ C$

Explanation:

Given that,

Area =7.70 cm²

Distance = 1.50 mm

Potential difference = 25.0 V

Suppose we find the electric field between the plates, the surface charge density, the capacitance and the charge on each plates.

We need to calculate the electric field

Using formula of electric field

$E=\dfrac{V}{d}$

Put the value into the formula

$E=\dfrac{25.0}{1.50\times10^{-3}}$

$E=16666.66\ V/m$

We need to calculate the charge density

Using formula of charge density

$\sigma=E\times\epsilon_{0}$

Put the value into the formula

$\sigma=16666.66\times8.85\times10^{-12}$

$\sigma=1.474\times10^{-7}\ C/m^2$

We need to calculate the capacitance

Using formula of capacitance

$C=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C=\dfrac{8.85\times10^{-12}\times7.60\times10^{-4}}{1.50\times10^{-3}}$

$C=4.484\times10^{-12}\ F$

We need to calculate the charge

Using formula of charge

$q=CV$

Put the value into the formula

$q=4.484\times10^{-12}\times25.0$

$q=112.1\times10^{-12}\ C$

Hence, The electric field is 16666.66 V/m.

The surface charge density is $1.474\times10^{-7}\ C/m^2$

The capacitance is $4.484\times10^{-12}\ F$

The charge on each plate is $112.1\times10^{-12}\ C$

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3 years ago

Two 1.0 kg masses are 4.0 m apart on a frictionless table. Each has 1.0μC of charge. Part A What is the magnitude of the electri

9966 [12]

Force between two charges is given by

$F =\frac{kq_1q_2}{r^2}$

$F =\frac{9*10^9* 1* 10^{-6}* 1 * 10^{-6}}{4^2}$

$F = 5.625 * 10^{-4} N$

Now in order to find the acceleration of each mass

we can use

F = ma

$5.625 * 10^{-4} = 1 * a$

$a= 5.625 * 10^{-4} m/s^2$

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4 years ago

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