What is the net force? 361 N m = 45 kg 112N 441 N

Tatiana [17]

Yes energy makes things move

6 0

4 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.

5 0

3 years ago

A point charge with a charge q1 = 4.00 μC is held stationary at the origin. A second point charge with a charge q2 = -4.10 μC mo

GrogVix [38]

Answer:

W = -0.480 J

Explanation:

given,

q₁ = 4 μC

q₂ = -4.10 μC

$W = kq_1q_2(\dfrac{1}{a}+\dfrac{1}{b})$

$b = \sqrt{(0.27-0)^2+(0.27-0)^2}$

b = 0.381

k = 8.99 × 10⁹ Nm²/C²

$W = 8.99\times 10^9\times 4\times 10^{-6}\times (-4.1 \times 10^{-6})(\dfrac{1}{0.17}+\dfrac{1}{0.381})$

$W = [-147.436\times (5.88-2.62)\times 10^{-3}]J$

W = -0.480 J

Work done by the electric force W = -0.480 J

4 0

3 years ago

What happens when an electric charge passes through a circuit?

sleet_krkn [62]

Answer:

The electric field direction within a circuit is by definition the direction that positive test charges are pushed. Thus, these negatively charged electrons move in the direction opposite the electric field.

Explanation:

: )

5 0

3 years ago

Read 2 more answers

What percentage of the intensity gets through both polarizers?

sp2606 [1]

Answer:

if the two polarizers have the same direction the transmitted light is 50% of the incident and if the two polarizers are at 90º the transmitted light is zero

Explanation:

The incident light is generally random, that is, it does not have a polarization plane, when the first polarized stops by half, this already polarized light arrives at the second polarizer and the causticity passes

I = I₀ cos² θ

therefore if the two polarizers have the same direction the transmitted light is 50% of the incident and if the two polarizers are at 90º the transmitted light is zero

8 0

3 years ago