15. Estimate the resistance ? of a standard (not a high-speed) phone charging cable.
?=_____Ω
19. A certain single ion channel is selectively permeable to K+ and has a resistance of 1.15 GΩ , therefore a current of potassium ions through the channel is possible. During an experiment, the channel is open for approximately 1.50 ms while the voltage across the channel is maintained at 89.5 mV . How many K+ ions travel through the channel during this time?
number of K+ ions:
21. Find the equivalent resistance of the combination of resistors ?1=42.0 Ω , ?2=75.0 Ω , ?3=33.0 Ω , ?4=61.0 Ω , ?5=12.5 Ω , and ?6=33.0 Ω , as shown in the figure.
equivalent resistance: ________Ω
22. The four resistors in the figure have an equivalent resistance of 16Ω. The resistances are as follows: ?1=12Ω, ?2=6.0Ω, and ?3=2.0Ω. Calculate the value of ?x.
?x=_______Ω
34. You charge an initially uncharged 78.7-mF capacitor through a 36.9-Ω resistor by means of a 9.00-V battery having negligible internal resistance. Find the time constant ? of the circuit.
?=________ s
What is the charge ? on the capacitor 1.49 time constants after the circuit is closed?
?=_______C
What is the charge ?0 after a long amount of time has passed?
?0=________C
15. Estimate the resistance ? of a standard (not a high-speed) phone charging cable. ?=_____Ω 19....
A certain single ion channel is selectively permeable to K and has a resistance of 1.15 GΩ. During an experiment, the channel is open for approximately 1.50 ms while the voltage across the channel is maintained at 81.5 mV with a patch clamp. How many K ions travel through the channel during this time?
Please help and show work. I keep getting the wrong answers for these problems 1. Cell membranes contain channels that allow Na + ions to rush in. Consider a channel that has a diameter of 2.91 nm and a length of 13.5 nm. If the channel has a resistance of 18.5 GΩ, what is the resistivity ? of the solution in the channel? 2. What is the resistance of a 54.154.1-m-long aluminum wire that has a diameter of 8.79 mm8.79...