The D on a violin has a frequency of 293 Hz when it is in tune. Calculate the period.
1 answer:
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The answer to this question is: it depends. It depends on the arrangement of the capacitors in a circuit: it can be either in series or in parallel. The difference is shown in the picture.
Capacitors are like batteries in a way that they store power from the source. It has some rules depending on the type of circuit. For parallel circuits, the voltage across each capacitor is equal. Therefore, V₁=V₂=V₃.
On the other hand, if the capacitors are arranged in series, the voltage across each capacitor should add up to the total voltage of the source. Therefore, V₁+V₂+V₃ = Total Voltage.
Capacitors are like batteries in a way that they store power from the source. It has some rules depending on the type of circuit. For parallel circuits, the voltage across each capacitor is equal. Therefore, V₁=V₂=V₃.
On the other hand, if the capacitors are arranged in series, the voltage across each capacitor should add up to the total voltage of the source. Therefore, V₁+V₂+V₃ = Total Voltage.
Answer:
K = 80.75 MeV
Explanation:
To calculate the kinetic energy of the antiproton we need to use conservation of energy:
<em>where : is the photon energy,
: are the rest energies of the proton and the antiproton, respectively, equals to m₀c²,
: are the kinetic energies of the proton and the antiproton, respectively, c: speed of light, and m₀: rest mass.</em>
Therefore the kinetic energy of the antiproton is:
<u>The proton mass is equal to the antiproton mass, so</u>:
Hence, the kinetic energy of the antiproton is 80.75 MeV.
I hope it helps you!
Answer:
A) Concentration of A left at equilibrium of we started the reaction with [A] = 2.00 M and [B] = 2.00 M is 0.55 M.
B) Final concentration of D at equilibrium if the initial concentrations are [A] = 1.00 M and [B] = 2.00 M is 0.90 M.
[D] = 0.90 M
Explanation:
With the first assumption that the volume of reacting mixture doesn't change throughout the reaction.
This allows us to use concentration in mol/L interchangeably with number of moles in stoichiometric calculations.
- The first attached image contains the correct question.
- The solution to part A is presented in the second attached image.
- The solution to part B is presented in the third attached image.
