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seen Jul 16 at 11:56

Oct
18
awarded  Revival
Oct
15
awarded  Yearling
Dec
28
revised What determines the pitch of a resonant object
added 47 characters in body
Dec
28
answered What determines the pitch of a resonant object
Dec
22
comment Does clipping to $m$ guarantee a maximum peak-to-peak amplitude $m$?
Ear damage is the result of accumulated exposure, that is, it depends on the total time the ear is exposed to high sound power levels. For example, damage to the ear may begin with 2 hours of exposure to 100dB at a loud concert, and builds up with say 8 hours of exposure to 85dB at work from using power tools. The damage to the ear is cumulative over a lifetime.
Dec
22
revised Does clipping to $m$ guarantee a maximum peak-to-peak amplitude $m$?
added 80 characters in body
Dec
22
answered Does clipping to $m$ guarantee a maximum peak-to-peak amplitude $m$?
Dec
22
answered Electric potential vs potential difference
Dec
22
revised separation between slip rings of DC generator
added 112 characters in body
Dec
22
comment separation between slip rings of DC generator
With two slip rings instead of a split ring, the direction of the current through the circuit will be reamain the same (a-b-c-d), but the rotor will have moved $90^{O}$. When the rotor moves past the $90^{O}$ point, the torque on the rotor will be in the opposite direction, unless we reverse the voltage/current, which is what the split ring does. The same happens after another $180^{O}$.
Dec
22
revised separation between slip rings of DC generator
added 1594 characters in body
Dec
22
answered separation between slip rings of DC generator
Nov
21
revised Solving a rather unusual (diagonal) circuit
deleted 1 characters in body
Nov
21
comment Solving a rather unusual (diagonal) circuit
The arrow on $U_{q3}$ is indicating that node F is 5V higher than node E. That is, the potential of F with respect to E is 5V, or $V_{FE}=5$. In the case of the current $I_{4}$ through resistor $R_{2}$, the voltage at A with respect to B is negative of the voltage of B with respect to A, where $V_{BA}=I_{4}R_{2}$. That is, if $I_{4}$ indicates positive current flow in the direction shown, the potential at B will be higher than the potential at A, so $V_{BA}$ will be positive.
Nov
21
comment Solving a circuit with Kirchoff/Ohms Law
Minor correction: the symbol for the battery is shown the wrong way around. The higher potential (+ve) terminal is the longer of the two parallel lines & must correspond with the arrow 'head' when indicating the polarity of $U_{0}$.
Nov
21
answered Solving a rather unusual (diagonal) circuit
Nov
18
revised Refractive index inside a fibre
added 698 characters in body
Nov
18
revised Refractive index inside a fibre
added 698 characters in body
Nov
18
answered Refractive index inside a fibre
Nov
18
awarded  Revival