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Komok [63]
4 years ago
9

A helium balloon has an internal pressure of 2.5 atm when it occupies 4.5 L. If it was compressed until it had a pressure of 3.3

atm, what would be the final volume? Report your answer with the correct number of significant figures.
Physics
2 answers:
steposvetlana [31]4 years ago
4 0

Answer: V = 3.4 L

Explanation: Use Boyle's Law to find the new volume. P1V1 = P2V2, derive for V2, then the formula will be V2= P1V1 / P2

V2 = 2.5 atm ( 4.5 L ) / 3.3 atm

= 3.4 L

kakasveta [241]4 years ago
4 0

Answer:3.40L

Explanation:

p1v1=p2v2

P1=2.5atm

V1= 4.5L

P2=3.3atm

V2=?

P1V1/P2=V2

V2=2.5*4.5/3.3

V2= 11.25/3.3

V2=3.40L

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Actually says Pantazis, since their frequencies are so wildly different, brain waves don’t interfere with radio waves. Even if that was the case, brain waves are so weak, they are hardly measurable at all. For comparison, says Pantazis, “the magnetic field of the earth is just strong enough to move the needle of a compass. Signals from the brain are a billionth of that strength.”
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4 years ago
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What are some of the phenotypic plant traits that scientists are investigating? Why do you think these traits were chosen?
vodomira [7]

The term phenotype in a more specific context to describe the collective expression of the genotype in conjunction with the environment on a plant's observable characteristics.

<u>Explanation</u>:

  • Phenotype, any of the measurable features of an organism arising from its genotype association (total genetic inheritance) with the environment. Behaviour, biochemical properties, colour, form, and scale are examples of measurable characteristics.
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8 0
3 years ago
For a proton in the ground state of a 1-dimensional infinite square well, what is the probability of finding the proton in the c
Butoxors [25]

Answer:

The probability of finding the proton at the central 2% of the well is almost exactly 4%

Explanation:

If we solve Schrödinger's equation for the infinite square well, we find that its eigenfunctions are sinusoidal functions, in particular, the ground state is a sinusoidal function for which only half a cycle fits inside the well.

let L be the well's length, the boundary conditions for the wavefunction are:

\psi(0) = \psi(L) =0

And Schrödinger's equation is:  

- \frac{\hbar^2}{2m} \frac{d^2\psi}{dx^2} = E\psi

The solution to this equation are sines and cosines, but the boundary conditions only allow for sine waves. As we pointed out, the ground state is the sine wave with the largest wavelength possible (that is, with the smallest energy).

\psi_0(x)=\sqrt[]{\frac{2}{L} }\, \sin(\frac{\pi x}{L} )\\

here the leading constant is just there to normalise the wavefunction.

Now, if we know the wavefunction, we can know what the probability density function is, it is:

f_X(x) = |\psi|^2

So in our case:

f_X(x) = \frac{2}{L} \sin^2(\frac{\pi x}{L})

And to find the probability of finding the particle in a strip at the centre of the well of width 2% of L we only have to integrate:

P(X \in [0.49 L, 0.51L ])= \int\limits^{0.49L}_{0.51L } {\frac{2}{L} \sin^2(\frac{\pi x}{L})} \, dx

If we do a substitution:

x = u \, L

We get the integral:

\int\limits^{0.49}_{0.51 } 2\,  \sin^2(\pi u)} \, du

This integral can be computed analytically, and it's numerical value is .0399868, that is, almost a 4% probability.

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