The answer is no. If you are dealing with a conservative force and the object begins and ends at the same potential then the work is zero, regardless of the distance travelled. This can be shown using the work-energy theorem which states that the work done by a force is equal to the change in kinetic energy of the object.
W=KEf−KEi
An example of this would be a mass moving on a frictionless curved track under the force of gravity.
The work done by the force of gravity in moving the objects in both case A and B is the same (=0, since the object begins and ends with zero velocity) but the object travels a much greater distance in case B, even though the force is constant in both cases.
Answer:
WB = 12 N
Explanation:
Solution:
- The relationship between wavelength (λ) and velocity (v) is given as:
v ∝ λ
- Both wavelength and and velocity are proportional. For the case B, we see that wavelength λ is halved. From using the above relation we can also say that velocity (v) is also halved.
- The relationship between velocity (v) and tension force (T) in the string is given by:
v^2 ∝ T
- We saw that if wavelength (λ) is halved, velocity (v) is halved then from the above relationship we see that Tension force T is reduced 4 times.
- Since the tension force T in part A is:
TA = m*g = WA = 48
- The Tension force in part B, TB is reduced 4 times as follows:
TB = 0.25*m*g = WB = 12 N
Complete Question
(A) Find the excitation energy from the ground level to the third excited level for an electron confined to a box that has a width of 0.125 nm .
(B) The electron makes a transition from the n=1 to n= 4 level by absorbing a photon. Calculate the wavelength of this photon.
Answer:
A
B
Explanation:
Considering question a
From the question we are told that
The width of the box is 
Generally the energy level of a particle confined to a box is mathematically represented as
Generally the excitation energy is mathematically represented as
![\Delta E = \frac{h^2 }{ 8 m L^2 } [n_2^2 - n_1 ^2 ]](https://tex.z-dn.net/?f=%5CDelta%20%20E%20%3D%20%20%5Cfrac%7Bh%5E2%20%7D%7B%208%20m%20L%5E2%20%7D%20%20%5Bn_2%5E2%20-%20n_1%20%5E2%20%20%5D)
From the question 
Here h is the Planck's constant with a value of 
m is the mass of electron with value 
So
![\Delta E = \frac{[6.62607015 * 10^{-34} ]^2 }{ 8 * (9.11 *10^{-31}) (0.125 *10^{-9})^2 } [3^2 - 1^2 ]](https://tex.z-dn.net/?f=%5CDelta%20%20E%20%3D%20%20%5Cfrac%7B%5B6.62607015%20%2A%2010%5E%7B-34%7D%20%5D%5E2%20%7D%7B%208%20%2A%20%289.11%20%2A10%5E%7B-31%7D%29%20%280.125%20%2A10%5E%7B-9%7D%29%5E2%20%7D%20%20%5B3%5E2%20-%201%5E2%20%20%5D)
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Considering question b
Generally the energy level of a particle confined to a box is mathematically represented as
Generally the excitation energy is mathematically represented as
From the question 
Here h is the Planck's constant with a value of
m is the mass of electron with value
So
![\Delta E = \frac{[6.62607015 * 10^{-34} ]^2 }{ 8 * (9.11 *10^{-31}) (0.125 *10^{-9})^2 } [4^2 - 1^2 ]](https://tex.z-dn.net/?f=%5CDelta%20%20E%20%3D%20%20%5Cfrac%7B%5B6.62607015%20%2A%2010%5E%7B-34%7D%20%5D%5E2%20%7D%7B%208%20%2A%20%289.11%20%2A10%5E%7B-31%7D%29%20%280.125%20%2A10%5E%7B-9%7D%29%5E2%20%7D%20%20%5B4%5E2%20-%201%5E2%20%20%5D)
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=> 
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Gnerally the wavelength is mathematically represented as
=> 
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Sound needs a medium to propagate through and this medium which is moves through is the air. The colder the temperature is, the less excited (meaning kinetic energy) the air particles will be meaning that the sound wave will travel slower. In the sahara desert, since the temperatures are warmer, and more excited, the sound will propagate faster through the air.
*Note that sound waves rely on vibration to move and if the air is hotter the molecules in the air are vibrate more than compared to lower temperatures.
