According to Newton’s second law of motion, force equals mass times acceleration (F = ma). Given that a car has a mass of 3,000
1 answer:
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A beat is an interference pattern between two sounds of slightly different frequencies, perceived as a periodic variation in volume whose rate is the difference of the two frequencies. Frequency beat is equal to,
The reference frequency in our case would be 392Hz, and since there is the possibility of the upper and lower range for the amount of beats per second that the two possible frequencies are heard would be
Therefore the two possible frequencies the piano wire is vibrating at, would be 396Hz and 388Hz
Answer:
1.1775 x 10^-3 m^3 /s
Explanation:
viscosity, η = 0.250 Ns/m^2
radius, r = 5 mm = 5 x 10^-3 m
length, l = 25 cm = 0.25 m
Pressure, P = 300 kPa = 300000 Pa
According to the Poisuellie's formula
Volume flow per unit time is
V = 1.1775 x 10^-3 m^3 /s
Thus, the volume of oil flowing per second is 1.1775 x 10^-3 m^3 /s.
Answer:
a) the capacitance is the same for both capacitors.
b) The potential difference between the plates for the capacitor with charge +2q, is double of the one for the capacitor with charge +q.
c) The electric field magnitude between the plates for the capacitor with charge +2q, is double of the one for the capacitor with charge +q
d) The energy stored between the plates for the capacitor with charge +2q, is 4 times the value for the one with charge +q
Explanation:
a) The capacitance of a capacitor, by definition, is as follows:
Appying Gauss' Law to one of plates, it can be showed, that the capacitance (for a parallel plates capacitor) can be expressed as follows:
C = ε*A / d
As it can be seen, it does not depend on the charge. so we conclude that the capacitance must be the same for both capacitors, due to they are identical except for the value of the charge on the plates.
b) By definition, as we said above, the capacitance is equal to the proportion between the charge of one of the plates, and the potential difference between them.
If this proportion must remain the same, and one of the capacitors has the double of the charge than the other, the potential difference must be the double also.
c) Applying Gauss' law, to the surface of one of the plates, and assuming a constant surface charge density σ, it can be showed that the electric field can be calculated as follows:
E*A = Q/ε₀ as σ=Q/A
⇒ E = σ/ε₀
As σ is directly proportional to the charge (being the area A the same), we conclude that the electric field for the capacitor with charge +2q must be the double than the one for the capacitior with charge +q.
c) The electric potential energy, stored between plates of a capacitor, can be written as follows:
If the capacitance remains the same, we can conclude that the electric potential energy for the capacitor with charge +2q, as the charge is raised to the 2nd power, must be 4 times the one for the capacitor with charge +q.