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3 years ago

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A gas occupies 200 mL at -73.0°C. To have the same gas occupy 300 mL, should the temperature be increased or decreased? What is

the new temperature in Kelvins? K

2 answers:

Tcecarenko [31]3 years ago

8 0

A gas occupies 200 mL at -73.0 degrees Celsius. To have the same gas occupy 300 milliliters, the temperature should be increased. 300K is the new temperature in Kelvins.

ValentinkaMS [17]3 years ago

7 0

<span>(PV)/T = (PV)/T. You can exclude the pressures since they are not mentioned. This means that V1/T1 = V2/T2. Kelvin is C + 273. Sooo, solving for T2=(V2T1)/V1. T2=(0.300L*200K)/0.200L=300K. NOTE: Volume should be in Liters, Temp in Kelvin, Pressure in atm (not used here).</span>

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A sphere of radius R contains charge Q spread uniformly throughout its volume. Find an expression for the electrostatic energy c

tensa zangetsu [6.8K]

Answer:

$E = \frac{3kQ^2}{5R}$

Explanation:

Let the sphere is uniformly charge to radius "r" and due to this charged sphere the electric potential on its surface is given as

$V = \frac{kq}{r}$

now we can say that

$q = \frac{Q}{\frac{4}{3}\pi R^3} (\frac{4}{3}\pi r^3)$

$q = \frac{Qr^3}{R^3}$

now electric potential is given as

$V = \frac{k\frac{Qr^3}{R^3}}{r}$

$V = \frac{kQr^2}{R^3}$

now work done to bring a small charge from infinite to the surface of this sphere is given as

$dW = V dq$

$dW = \frac{kQr^2}{R^3} dq$

here we know that

$dq = \frac{3Qr^2dr}{R^3}$

now the total energy of the sphere is given as

$E = \int dW$

$E = \int_0^R \frac{kQr^2}{R^3} (\frac{3Qr^2dr}{R^3})$

$E = \frac{3kQ^2}{R^6} (\frac{R^5}{5} - 0)$

$E = \frac{3kQ^2}{5R}$

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4 years ago

At room temperature, sound travels at a speed of about 344 m/s in air. You see a distant flash of lightning and hear the thunder

bearhunter [10]

Answer:

<em>The lighten travels 0.853 miles.</em>

Explanation:

Sound: Sound is a form of wave which is conveyed through an elastic medium from a vibrating body to a listener.

v = 2x/t .......................................... Equation 1

making x the subject of the equation

x = vt/2........................................ Equation 2

Where v = velocity of sound in air, x = distance traveled by the sound, t = time

Given: v = 344 m/s t = 8 s

Substituting into equation 2

x = 344(8)/2

x = 1376 m.

x = 1376×0.00062 miles = 0.853 miles

<em>Thus the lighten travels 0.853 miles.</em>

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3 years ago

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Ans is a male having the disease.

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3 years ago

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(a) If a proton with a kinetic energy of 6.2 MeV is traveling in a particle accelerator in a circular orbit with a radius of 0.5

Tju [1.3M]

Answer:

The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

Explanation:

Suppose Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$

Given that,

Kinetic energy = 6.2 MeV

Radius = 0.500 m

We need to calculate the acceleration

Using formula of acceleration

$a=\dfrac{v^2}{r}$

Put the value into the formula

$a=\dfrac{\dfrac{1}{2}mv^2}{\dfrac{1}{2}mr}$

Put the value into the formula

$a=\dfrac{6.2\times10^{6}\times1.6\times10^{-19}}{\dfrac{1}{2}\times1.67\times10^{-27}\times0.51}$

$a=2.32\times10^{15}\ m/s^2$

We need to calculate the rate at which it emits energy because of its acceleration is

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$

Put the value into the formula

$\dfrac{dE}{dt}=\dfrac{(1.6\times10^{-19})^2\times(2.3\times10^{15})^2}{6\pi\times8.85\times10^{-12}\times(3\times10^{8})^3}$

$\dfrac{dE}{dt}=3.00\times10^{-23}\ J/s$

The energy in ev/s

$\dfrac{dE}{dt}=\dfrac{3.00\times10^{-23}}{1.6\times10^{-19}}\ J/s$

$\dfrac{dE}{dt}=1.875\times10^{-4}\ ev/s$

We need to calculate the fraction of its energy that it radiates every second

$\dfrac{\dfrac{dE}{dt}}{E}=\dfrac{1.875\times10^{-4}}{6.2\times10^{6}}$

$\dfrac{\dfrac{dE}{dt}}{E}=3.02\times10^{-11}$

Hence, The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

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3 years ago

An object accelerates to a velocity of 230 m/s over a time of 2.5 s. The acceleration it experienced was 42 m/s2. What was its i

zmey [24]

Answer:

230 = x + 105

x= 125

Explanation:

v = v0 + at

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3 years ago