What happens to the resistance of a wire as it gets wet

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

Lester plays middle linebacker for dices football team. During one play, he made the following movements. First, he back-pedaled

NikAS [45]

Answer:

Explanation:

missing info .....

4 0

2 years ago

The chart show the masses and velocities of two colliding objects that stick together after a collision.

finlep [7]

Answer:

1,500 kg.m/s

Explanation:

First, arrange the information in a table:

Object Mass (kg) Velocity (m/s)

A 200 15

B 150 - 10

After the collision, the two objects are stick together, thus you talk aobut one object and one momentum.

According to the law of convervation of momentum, the momentum after the collision is equal to the momentum before the collision.

Momentum before the collision, P₁:

$P_1=m_{A,1}\times v_{A,1}+m_{B,1}\times v_{B,1}$

$P_1=200kg\times 15m/s+150kg\times (-10m/s)\\\\ P_1=3000kg.m/s-1500kg.m/s=1500kg.m/s$

Momentum after the collision:

As stated, it es equal to the momentum before the collision: 1,500 kg . m/s

6 0

4 years ago

Two football players run towards each other along a straight path in Penrith Park in the clash between the Melbourne storms and

Agata [3.3K]

Answer:

a) v = 0.4799 m / s, b) K₀ = 1600.92 J, K_f = 5.46 J

Explanation:

a) How the two players collide this is a momentum conservation exercise. Let's define a system formed by the two players, so that the forces during the collision are internal and also the system is isolated, so the moment is conserved.

Initial instant. Before the crash

p₀ = m v₁ + M v₂

where m = 95 kg and his velocity is v₁ = -3.75 m / s, the other player's data is M = 111 kg with velocity v₂ = 4.10 m / s, we have selected the direction of this player as positive

Final moment. After the crash

p_f = (m + M) v

as the system is isolated, the moment is preserved

p₀ = p_f

m v₁ + M v₂ = (m + M) v

v = $\frac{m v_1 + M v_2}{m+M}$

let's calculate

v = $\frac{ -95 \ 3.75 \ + 111 \ 4.10}{95+111}$

v = 0.4799 m / s

b) let's find the initial kinetic energy of the system

K₀ = ½ m v1 ^ 2 + ½ M v2 ^ 2

K₀ = ½ 95 3.75 ^ 2 + ½ 111 4.10 ^ 2

K₀ = 1600.92 J

the final kinetic energy

K_f = ½ (m + M) v ^ 2

k_f = ½ (95 + 111) 0.4799 ^ 2

K_f = 5.46 J

3 0

3 years ago

The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the t

DENIUS [597]

Answer:

terminal velocity is;

v = 117.54 m/s

v = 423.144 km/hr

Explanation:

Given the data in the question;

we know that, the force on a body due to gravity is;

$F_g$ = mg

where m is mass and g is acceleration due to gravity

Force of drag is;

$F_d$ = $\frac{1}{2}$ pCAv²

where p is the density of fluid, C is the drag coefficient, A is the area and v is the terminal velocity.

Terminal velocity is reach when the force of gravity is equal to the force of drag.

$F_g = F_d$

mg = $\frac{1}{2}$ pCAv²

we solve for v

v = √( 2mg / pCA )

so we substitute in our values

v = √( [2×(86 kg)×9.8 m/s² ] / [ 1.21 kg/m³ × 0.7 × 0.145 m²] )

v = √( 1685.6 / 0.122015 )

v = √( 13814.6949 )

v = 117.54 m/s

v = ( 117.54 m/s × 3.6 ) = 423.144 km/hr

Therefore terminal velocity is;

v = 117.54 m/s

v = 423.144 km/hr

5 0

3 years ago