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3 years ago

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If the temperature of a volume of ideal gas increases for 100°C to 200°C, what happens to the average kinetic energy of the mole

cules?

1 answer:

mina [271]3 years ago

8 0

Answer: $\Delta E=2.0715\times 10^{-21} J$

Explanation:

Given

Temperature of the gas is increased from 100 to 200

Also we know that average kinetic energy of the molecules is

$E=\frac{3}{2}\cdot \frac{R}{N_A}T$

Where

R=Gas constant

$N_A$ =Avogadro's number

T=Temperature in kelvin

$\frac{R}{N_A}=1.381\times 10^{-23}$

So kinetic energy increases by

$\Delta E=\frac{3}{2}\times 1.381\times 10^{-23}\left ( 200-100\right )$

$\Delta E=2.0715\times 10^{-21} J$

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What are the variables in Gay-Lussac's law? pressure and volume pressure, temperature, and volume pressure and temperature volum

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pressure and temperature (assuming volume is constant)

Explanation:

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Careful measurements have been made of Olympic sprinters in the 100-meter dash. A quite realistic model is that the sprinter's v

mihalych1998 [28]

Answer:

a.

$\displaystyle a(0 )=8.133\ m/s^2$

$\displaystyle a(2)=2.05\ m/s^2$

$\displaystyle a(4)=0.52\ m/s^2$

b. $\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15$

c. $t=9.9 \ sec$

Explanation:

Modeling With Functions

Careful measurements have produced a model of one sprinter's velocity at a given t, and it's is given by

$\displaystyle V(t)=a(1-e^{bt})$

For Carl Lewis's run at the 1987 World Championships, the values of a and b are

$\displaystyle a=11.81\ ,\ b=-0.6887$

Please note we changed the value of b to negative to make the model have sense. Thus, the equation for the velocity is

$\displaystyle V(t)=11.81(1-e^{-0.6887t})$

a. What was Lewis's acceleration at t = 0 s, 2.00 s, and 4.00 s?

To compute the accelerations, we must find the function for a as the derivative of v

$\displaystyle a(t)=\frac{dv}{dt}=11.81(0.6887\ e^{0.6887t})$

$\displaystyle a(t)=8.133547\ e^{-0.6887t}$

For t=0

$\displaystyle a(0)=8.133547\ e^o$

$\displaystyle a(0 )=8.133\ m/s^2$

For t=2

$\displaystyle a(2)=8.133547\ e^{-0.6887\times 2}$

$\displaystyle a(2)=2.05\ m/s^2$

$\displaystyle a(4)=8.133547\ e^{-0.6887\times 4}$

$\displaystyle a(4)=0.52\ m/s^2$

b. Find an expression for the distance traveled at time t.

The distance is the integral of the velocity, thus

$\displaystyle X(t)=\int v(t)dt \int 11.81(1-e^{-0.6887t})dt=11.81(t+\frac{e^{-0.6887t}}{0.6887})+C$

$\displaystyle X(t)=11.81(t+1.45201\ e^{-0.6887t})+C$

To find the value of C, we set X(0)=0, the sprinter starts from the origin of coordinates

$\displaystyle x(0)=0=>11.81\times1.45201+C=0$

Solving for C

$\displaystyle c=-17.1482\approx -17.15$

Now we complete the equation for the distance

$\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15$

c. Find the time Lewis needed to sprint 100.0 m.

The equation for the distance cannot be solved by algebraic procedures, but we can use approximations until we find a close value.

We are required to find the time at which the distance is 100 m, thus

$\displaystyle X(t)=100=>11.81(t+1.45\ e^{-0.6887t})-17.15=100$

Rearranging

$\displaystyle t+1.45\ e^{-0.6887t}=9.92$

We define an auxiliary function f(t) to help us find the value of t.

$\displaystyle f(t)=t+1.45\ e^{-0.687t}-9.92$

Let's try for t=9 sec

$\displaystyle f(9)=9+1.45\ e^{-0.687\times 9}-9.92=-0.92$

Now with t=9.9 sec

$\displaystyle f(9.9)=9.9+1.45\ e^{-0.687\times 9.9}-9.92=-0.0184$

That was a real close guess. One more to be sure for t=10 sec

$\displaystyle f(10)=10+1.45\ e^{-0.687\times 10}-9.92=0.081$

The change of sign tells us we are close enough to the solution. We choose the time that produces a smaller magnitude for f(t).

$At t\approx 9.9\ sec, \text{ Lewis sprinted 100 m}$

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What does the "coefficient of friction" tell you?

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a coefficient of fraction is a value that shows the relationship between two objects and the normal reaction between the objects that are involved.

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The depth of the Pacific Ocean in the Mariana Trench is 36,198 ft. What is the gauge pressure at this depth

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Answer:

the pressure at the depth is 1.08 × $10^{8}$ Pa

Explanation:

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P = h $\rho$ g

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$\rho$ = density of water = 1000 $\frac{kg}{m^{3} }$

g = acceleration due to gravity ≈ 9.8 $\frac{m}{s^{2} }$

P = 11033.15 × 9.8 × 1000

P = 1.08 × $10^{8}$ Pa

Thus, the pressure at the depth is 1.08 × $10^{8}$ Pa

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3 years ago

I just need x isolated

mamaluj [8]

Answer:

$x=\frac{-y+\sqrt{y^2+4rt} }{2r}$

$x=\frac{-y-\sqrt{y^2+4rt} }{2r}$

Explanation:

$rx+y=\frac{t}{x}\\\\x(rx+y)=(\frac{t}{x})x\\\\rx^2+yx=t\\\\rx^2+yx-t=t-t\\\\rx^2+yx-t=0$

Solve using the quadratic formula.

$x=\frac{-y+\sqrt{y^2+4rt} }{2r}$

$x=\frac{-y-\sqrt{y^2+4rt} }{2r}$

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