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VLD [36.1K]
3 years ago
8

A gas sample is heated from -20.0°C to 57.0°C and the volume is increased from 2.00 L to 4.50 L. If the initial pressure is 0.14

0 atm, what is the final pressure? Select one: a. 0.0477 atm b. –0.177 atm c. 0.411 atm d. 0.242 atm e. 0.0811 atm
Physics
1 answer:
Svetach [21]3 years ago
5 0

Answer:

The answer is e.

Explanation:

We take:

T_{1}=253K

V_{1}=2.00 l

P_{1}=0.14atm

V_{2}=4.50 l

T_{2}=330K

Taking the gas as an ideal gas, we can use the ideal gas law:

\frac{PV}{nRT}

⇒ n=\frac{P_{1}V_{1}}{RT_{1}} ⇒ n=0.013mol

Then:

P_{2}=\frac{nRT_{2}}{V_{2}} ⇒ P_{2}=0.0811 atm

Taking R=0.08205atmL/molK.

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I am going to say

C.  Energy contained in the nucleus of an atom

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4 years ago
In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 1
asambeis [7]

Explanation:

It is given that,

Frequency of monochromatic light, f=5\times 10^{14}\ Hz

Separation between slits, d=2.2\times 10^{-5}\ m

(a) The condition for maxima is given by :

d\ sin\theta=n\lambda

For third maxima,

\theta=sin^{-1}(\dfrac{n\lambda}{d})

\theta=sin^{-1}(\dfrac{n\lambda}{d})

\theta=sin^{-1}(\dfrac{nc}{fd})  

\theta=sin^{-1}(\dfrac{3\times 3\times 10^8\ m/s}{5\times 10^{14}\ Hz\times 2.2\times 10^{-5}\ m})  

\theta=4.69^{\circ}

(b) For second dark fringe, n = 2

d\ sin\theta=(n+1/2)\lambda

\theta=sin^{-1}(\dfrac{5\lambda}{2d})

\theta=sin^{-1}(\dfrac{5c}{2df})

\theta=sin^{-1}(\dfrac{5\times 3\times 10^8}{2\times 2.2\times 10^{-5}\times 5\times 10^{14}})

\theta=3.90^{\circ}

Hence, this is the required solution.

8 0
3 years ago
A mass of 3.6 kg oscillate on a horizontal spring with a spring constant of 160 N/m.
Darya [45]

Answer:

48.7 J

Explanation:

For a mass-spring system, there is a continuous conversion of energy between elastic potential energy and kinetic energy.

In particular:

- The elastic potential energy is maximum when the system is at its maximum displacement

- The kinetic energy is maximum when the system passes through the equilibrium position

Therefore, the maximum kinetic energy of the system is given by:

KE=\frac{1}{2}mv^2

where

m is the mass

v is the speed at equilibrium position

In this problem:

m = 3.6 kg

v = 5.2 m/s

Therefore, the maximum kinetic energy is:

KE=\frac{1}{2}(3.6)(5.2)^2=48.7 J

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3 years ago
Uranium has two naturally occurring isotopes. 238u has a natural abundance of 99.3% and 235u has an abundance of 0.7%. it is the
dmitriy555 [2]

The ration of the rms speed of 235uf6 to that of 238uf6 is 1.004.

The molecular mass of 235uf6 is 349, while that of 238uf6 is 352.

The rms speed is calculated as

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Thus the ratio rms speed of 235uf6 to 238uf6 is calculated as

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6 0
3 years ago
The upward velocity of a 2540kg rocket is v(t)=At + Bt2. At t=0 a=1.50m/s2. The rocket takes off and one second afterwards v=2.0
Alexeev081 [22]

Answer:

The value of A is 1.5m/s^2 and B is 0.5m/s^³

Explanation:

The mass of the rocket = 2540 kg.

Given velocity, v(t)=At + Bt^2

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a= 1.50 m/s^2

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A = 1.5m/s^2

now,

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1.5× 1 +B× 1 = 2m/s

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B = 0.5m/s^³

Now Check V(t) = A× t + B × t^²

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7 0
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