All of them are examples of energy being transferred by waves<span />
Answer: If both gases undergo the same entropy then more heat is added to gas a because the entropy of the gas a is less than the entropy of the gas b.
Explanation:
Entropy is defined as the degree of randomness. When the temperature of the gas increases then the entropy of gas also increases.
In the given problem, Quantity a of an ideal gas is at absolute temperature t, and a second quantity b of the same gas is at absolute temperature 2t.
Heat is added to each gas, and both gases are allowed to expand isothermally. It means that the volume is constant during this process.
If both gases undergo the same entropy then more heat is added to gas a because the entropy of the gas a is less than the entropy of the gas b. If the heat is added then there will be more entropy.
Answer:
Coulomb's law is:
First, force has units of Newtons, the charges have units of Coulombs, and r, the distance, has units of meters, then, working only with the units we have:
N = (1/{e0})*C^2/m^2
then we have:
{e0} = C^2/(m^2*N)
And we know that N = kg*m/s^2
then the dimensions of e0 are:
{e0} = C^2*s^2/(m^3)
(current square per time square over cubed distance)
And knowing that a Faraday is:
F = C^2*S^2/m^2
The units of e0 are:
{e0} = F/m.
(a) The proton’s potential energy change is 3.6 x 10⁻¹⁸ J.
(b) The potential difference between the negative plate and a point midway between the plates is 11.25 V.
(c) The speed of the proton just before it hits the negative plate is 6.57 x 10⁴ m/s.
<h3>
Potential energy of the proton</h3>
U = qΔV
where;
- q is charge of the proton
- ΔV is potential difference
U = q(Ed)
U = (1.6 x 10⁻¹⁹)(1500 x 1.5 x 10⁻²)
U = 3.6 x 10⁻¹⁸ J
<h3>Potential difference between the negative plate and a point midway</h3>
ΔV = E(0.5d)
ΔV = 0.5Ed
ΔV = 0.5 (1500)(1.5 x 10⁻²)
ΔV = 11.25 V
<h3>Speed of the proton </h3>
U = ¹/₂mv²
U = mv²
v² = 2U/m
where;
- m is mass of proton = 1.67 x 10⁻²⁷ kg
v² = (2 x 3.6 x 10⁻¹⁸) / ( 1.67 x 10⁻²⁷)
v² = 4.311 x 10⁹
v = √(4.311 x 10⁹)
v = 6.57 x 10⁴ m/s
Thus, the proton’s potential energy change is 3.6 x 10⁻¹⁸ J.
The potential difference between the negative plate and a point midway between the plates is 11.25 V.
The speed of the proton just before it hits the negative plate is 6.57 x 10⁴ m/s.
Learn more about potential difference here: brainly.com/question/24142403
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The atomic number of Aluminum is 13. The atomic number represents the number of protons in an atom.
