In light of this, V=V 0 loge (r/r 0 ) Field E= dr dV =V 0(r0r) eE= r mV2 alternatively, reV0r0=rmV2. V=(m eV 0 r 0 ) \ s1 / 2mV=(m e V 0 r 0 ) 1/2 = constant mvr= 2 nh, also known as Bohr's quantum condition or Hermitian matrix.
Show that the eigenfunctions for the Hermitian matrix in review exercise 3a can be normalized and that they are orthogonal.
Demonstrate how the pair of degenerate eigenvalues for the Hermitian matrix in review exercise 3b can be made to have orthonormal eigenfunctions.
Under the given Hermitian matrix, "border conditions," solve the following second order linear differential equation: d2x/ dt2 + k2x(t) = 0 where x(t=0) = L and dx(t=0)/ dt = 0.
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Answer:
Option B. The planet is a very way from the center or the Sun.
Explanation:
If the surface temperature of the newly formed planet is - 20 K that means the distance of the planet from the sun in greater.
If we look at the temperature trends of the planets in our solar system, the planets closer to the sun like Mercury, Venus have very high surface temperature but as we move move away from the the center of the solar system, i.e., Sun, the temperature goes on decreasing.
You are asked to calculate an object's velocity.
In order to do so, you must know the object's speed and the direction
in which it's moving, or you must have other information that makes it
possible for you to calculate its speed and direction.
Answer: Its the last one: The position of the block will change
Explanation: Since the forces are unbalanced, one side of the block will be stronger, therefore making the block move.
Answer:
With this information is not possible to calculate the mass.
Explanation:
This is a characteristic problem of energy conservation, where kinetic energy becomes potential energy. For this particular problem, we have the initial speed as input data. The moment the ball comes out of the cannon we have the maximum kinetic energy, as the ball goes up the ball will gain more potential energy as the ball loses kinetic energy, until the moment the ball reaches the maximum height. At the maximum height point, the ball will have its maximum potential energy while its kinetic energy is zero. In other words, all the kinetic energy that was, in the beginning, was transformed into potential energy.
In the above equation the masses are canceled and we can determine the maximum height, by means of the initial speed.
![h=\frac{0.5*v^2}{g} [m]](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B0.5%2Av%5E2%7D%7Bg%7D%20%5Bm%5D)
But the mass cannot be determined, since it would be necessary to know the value of the energy, in order to determine the value of the mass.
