Answer:
I think when the object transferring the thermal energy reaches the same temp as the object absorbing it
Explanation:
An electron cloud represents all the orbitals in an atom.
Given Information:
KEa = 9520 eV
KEb = 7060 eV
Electric potential = Va = -55 V
Electric potential = Vb = +27 V
Required Information:
Charge of the particle = q = ?
Answer:
Charge of the particle = +4.8x10⁻¹⁸ C
Explanation:
From the law of conservation of energy, we have
ΔKE = -qΔV
KEb - KEa = -q(Vb - Va)
-q = KEb - KEa/Vb - Va
-q = 7060 - 9520/27 - (-55)
-q = 7060 - 9520/27 + 55
-q = -2460/82
minus sign cancels out
q = 2460/82
Convert eV into Joules by multiplying it with 1.60x10⁻¹⁹
q = 2460(1.60x10⁻¹⁹)/82
q = +4.8x10⁻¹⁸ C
The coefficient of linear expansion, given that the length of the pipe increased by 1.5 cm is 1.67×10¯⁵ /°F
<h3>How to determine the coefficient of linear expansion</h3>
From the question given above, the following data were obtained
- Original diameter (L₁) = 10 m
- Change in length (∆L) = 1.5 cm = 1.5 / 100 = 0.015 m
- Change in temperature (∆T) = 90 °F
- Coefficient of linear expansion (α) =?
The coefficient of linear expansion can be obtained as illustrated below:
α = ∆L / L₁∆T
α = 0.015 / (10 × 90)
α = 0.015 / 900
α = 1.67×10¯⁵ /°F
Thus, we can conclude that the coefficient of linear expansion is 1.67×10¯⁵ /°F
Learn more about coefficient of linear expansion:
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Answer:
The answer to this question can be defined as follows:
Explanation:
Therefore the 4th harmonicas its node is right and over the pickup so, can not be captured from 16.25, which is 1:4 out of 65. Normally, it's only conceptual for the certain harmonic, this will be low, would still be heard by the catcher.
Instead, every harmonic node has maximum fractions along its string; the very first node is the complete string length and the second node is half a mile to the third node, which is one-third up and so on.
