A stretched string has a mass per unit length of 5.12 g/cm and a tension of 19.3 N. A sinusoidal wave on this string has an ampl
1 answer:
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Answer:
<u>For proton:</u>
A. The proton is released from Vb (highest potential)
B. v = 2.9x10⁴ m/s
<u>For electron:</u>
A. The electron is released from Va (lowest potential)
B. v = 1.2x10⁶ m/s
Explanation:
<u>For a proton we have</u>:
A. To find the origin from which the proton was released we need to remember that in a potential difference, a proton moves from the highest potential to the lowest potential.
Having that:
Va = 1.51 V and Vb = 5.81 V
We can see that the proton moves from Vb to Va, hence the proton was released from Vb.
B. We now that the work done by an electric field is given by:
(1)
Where:
q: is the proton's charge = 1.6x10⁻¹⁹ C
V: is the potential
Also, the work is equal to:
(2)
Where:
K: is the kinetic energy
m: is the proton's mass = 1.67x10⁻²⁷ kg
: is the velocity in the point a
: is the velocity in the point b = 0 (starts from rest)
Matching equation (1) with (2) we have:
<u>For an electron we have</u>:
A. For an electron we know that it moves from the lowest potential (Va) to the highest potential (Vb), so it is released from Va.
B. The speed is:
Since = 0 (starts from rest) and
= 9.1x10⁻³¹ kg (electron's mass), we have:
I hope it helps you!
Answer:
The net work produced is 30.37 KJ
The efficiency of cycle is 72.3%
Explanation:
For the net work produced, we have the formula:
Work = (Th - Tc)(Sh - Sc)(M)
Where,
Th = higher temperature = 800° C + 273 = 1073 k
Tc = lower temperature = 25° C + 273 = 298 k
Sh = specific entropy at higher temperature
Sc = specific entropy at lower temperature
M = molar mass of air
Using, ideal gas table to find entropy. The table is attached.
therefore,
Work = (1073 k - 298 k)(3.0485235 KJ/kg.k - 1.69528 KJ/kg.k)(0.02896 kg)
Work = (775 k)(1.3532435 KJ/kg.k)(0.02896 kg)
<u>Work = 30.37 KJ</u>
Now, for the efficiency (n), we have a formula:
n = 1 - Tc/Th
n = 1 - (298 k)/(1073 k)
<u>n = 0.723 = 72.3 %</u>
