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vladimir1956 [14]
2 years ago
10

The first spacecraft which did not merely fly by a jovian (or giant) planet, but actually went into orbit around it for an exten

ded period of time was
Physics
1 answer:
Alenkinab [10]2 years ago
3 0

Answer:    Four spacecraft (Pioneer 10 & 11, then Voyager 1 & 2) had previously flown by the Jupiter system, but the Galileo mission was the first to enter orbit around the planet.

Explanation:

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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 ball with mass of 0.050 kg is dropped from a height of h1 = 1 .5 m. It collides with the floor, then bounces up to a height of
Iteru [2.4K]

Answer:

Explanation:

Impulse of reaction force of floor = change in momentum

Velocity of impact = √ 2gh₁

= √ 2 x 9.8 x 1.5 = 5.4 m /s.

velocity of rebound = √2gh₂

= √ 2x 9.8 x 1

= 4.427 m / s.

Initial momentum = .050 x 5.4 = .27 kg m/s

Final momentum = .05 x 4.427 = .22 kg.m/s

change in momentum = .27 - .22 = .05 kg m/s

Impulse = .05 kg m /s

Impulse = force x time

force = impulse / time

.05 / .015 = 3.33 N.

kinetic energy = 1/2 m v²

Initial kinetic energy = 1/2 x .05 x 5.4²

= 0.729 J

Final Kinetic Energy =1/2 x .05 x 4.427²

= 0.489 J

Change in Kinetic energy =0 .24 J

Lost kinetic energy is due to conversion of energy into sound light etc.

4 0
4 years ago
What is the relationship between distance of the objects and the gravitational force between them? (Another way of asking: What
Nadya [2.5K]

Explanation:

increase in the force of gravity

6 0
3 years ago
Describe the energy transformations that occur when you bounce a ball?
Gemiola [76]
The transformation is kinetic energy
7 0
4 years ago
Read 2 more answers
A stone has a mass of 100 grams and a volume of 10ml water has a density of 1 g/ml will the stone sink or float?
JulsSmile [24]
Density = Mass / Volume 

Density of stone = 100 / 10 = 10 g/mL

If the density is higher than water, then the object will sink. If the density of the object is lower than water, then the object will float. 

Since the density of the rock is 10g/mL and the density of the water is 1g/mL, the stone will sink.
3 0
3 years ago
Read 2 more answers
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