<h2>The work done = - 2 x 10⁴ J</h2>
Explanation:
In the first case , the volume is kept constant and pressure varies .
In isothermal process , the work done
W₁ = V x ΔP
here V is the volume of gas and ΔP is the change in pressure
Thus W₁ = 0
Because there is no change in volume , therefore displacement is zero .
In second case pressure is constant , but volume changes
Thus W₂ = P x ΔV
here P is the pressure and ΔV is the change in volume
Therefore W₂ = 4 x 10⁵ x 5 x 10⁻² = 2 x 10⁴ J
The total work done W = - 2 x 10⁴ J
Because the work done in compression is negative .
Answer:
(a) The energy of the photon is 1.632 x 
J.
(b) The wavelength of the photon is 1.2 x 
m.
(c) The frequency of the photon is 2.47 x 
Hz.
Explanation:
Let;
= -13.60 ev
= -3.40 ev
(a) Energy of the emitted photon can be determined as;
- = -3.40 - (-13.60)
= -3.40 + 13.60
= 10.20 eV
= 10.20(1.6 x 
)
- = 1.632 x Joules
The energy of the emitted photon is 10.20 eV (or 1.632 x Joules).
(b) The wavelength, λ, can be determined as;
E = (hc)/ λ
where: E is the energy of the photon, h is the Planck's constant (6.6 x 
Js), c is the speed of light (3 x 
m/s) and λ is the wavelength.
10.20(1.6 x ) = (6.6 x * 3 x )/ λ
λ = 
= 1.213 x
Wavelength of the photon is 1.2 x m.
(c) The frequency can be determined by;
E = hf
where f is the frequency of the photon.
1.632 x = 6.6 x x f
f = 
= 2.47 x Hz
Frequency of the emitted photon is 2.47 x Hz.
Please provide the choices to select the possible choices.
The output waveforms after passing through the transformer actually depend on the type of transformer used. It could either be a step-up transformer (steps voltage up), or a step-down transformer (steps voltage down). Both transformers have an output voltage in a form of a sine wave.
Thank you for posting your question here. The total distance traveled by the truck during the 3.2 seconds interval is 83 m. Below is the solution:
d = vit + 1/2 at^2
d = (22m/ s) (3.2s) + 1/2 (2.4m/ s^2) (3.2s)^2
d = 83 m
Hope the answer helps.
