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Maksim231197 [3]
2 years ago
14

Let this oscillator have the same energy as a mass on a spring, with the same k and m, released from rest at a displacement of 5

.00cm from equilibrium. what is the quantum number n of the state of the harmonic oscillator? express the quantum number to three significant figures.
Physics
1 answer:
allochka39001 [22]2 years ago
3 0
There are a lot of same examples that you may have worked before, where the mass on a spring uses a classics when it comes to mechanics. So in this system, always put in your mind that there is an enormous quantum standard that one can use in the equation. It should be 2.10x10 raise to a negative sixth. J.
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Answer: The energy delivered to the toaster is 264.490KJ

Explanation:

Here is the complete question:

The resistance of a bagel toaster is 14 ?. To prepare a bagel, the toaster is operated for one minute from a 120-V outlet. How much energy is delivered to the toaster?

Step-by-step explanation:

Please see attachment below

7 0
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Based on the direction of propagation compared to direction of vibration, waves are classified into:
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For the question we have here, since the direction of the wave is the same as the direction of vibration of particles, therefore, this wave is a longitudinal wave
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robby skateboards 0.50 blocks to his friend's house in 1.2 minutes. what is his speed? 6.0 blocks/min 6.0 blocks/min, in the dir
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Calculate the mass of air in a room of floor dimensions =10M×12M and height 4m(Density of air =1.26kg/m cubic​
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The volume of the room is the product of its dimensions:

10\times 12 \times 4 = 480\text{ m}^3

Now, from the equation

d=\dfrac{m}{V}

where d is the density, m is the mass and V is the volume, we deduce

m=dV

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Two flat conductors are placed with their inner faces separated by 6.0 mm. If the surface charge density on one of the inner fac
dangina [55]

Explanation:

Relation between electric field and charge density is as follows.

           E = \frac{\sigma}{2 \epsilon}

where,    \sigma = charge density

              \epsilon = permittivity of free space = 8.85 \times 10^{-12}

So,  E_{\text{outside}} = 0

      E_{inside} = \frac{+\sigma}{2 \epsilon} - \frac{-\sigma}{2 \epsilon}

or,     E_{inside} = \frac{\sigma}{\epsilon}

Now, formula to calculate the potential difference of two conductors is as follows.

         V_{1} - V_{2} = \frac{\sigma \times d}{\epsilon}

It is given that,

           d = 6.0 mm = 6 \times 10^{-3} m

        \sigma = 40 \times 10^{-12} C/m^{2}

Hence, we will calculate the magnitude of the electric potential differences between the two conductors as follows.

        V_{1} - V_{2} = \frac{\sigma \times d}{\epsilon}

                     = \frac{40 \times 10^{-12} \times 6 \times 10^{-3}}{8.85 \times 10^{-12}}      

                     = 0.0271 volts

thus, we can conclude that value of the magnitude of the electric potential differences between the two conductors is 0.0271 volts.

7 0
2 years ago
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