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Strike441 [17]
4 years ago
11

Look at Concept Simulation 5.2 to review the concepts involved in this question. Two cars are identical, except for the type of

tread design on their tires. The cars are driven at the same speed and enter the same unbanked horizontal turn. Car A cannot negotiate the turn, but car B can. Which tread design, the one on car A or the one on car B, yields a larger coefficient of static friction between the tires and the road?
Physics
1 answer:
dimulka [17.4K]4 years ago
7 0

Answer:

Tread design on car B would yield a larger coefficient of static friction between the tires and the road

Explanation:

The car model using the coefficient of static friction doesn't work well with tires. A higher coefficient of static friction would require more force to cause a loss of attraction.

The static frictional force helps to keep the unbanked horizontal turn. This means that the frictional force is the centripetal force.

The tread design of car B ensures that the centripetal force is enough to negotiate the turn. On the other hand, the tread design of car A does not provide the necessary centripetal force, hence car A is unable to negotiate the turn.

Therefore, tread design on car B would yield a larger coefficient of static friction between the tires and the road.

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Nerve cells transmit electric signals through their long tubular axons. These signals propagate due to a sudden rush of Na+ ions
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Answer:

I = \frac{9.12x10^{-8} C}{0.007 s}= 1.30285 x10^{-5} \frac{C}{s}=1.30285 x10^{-5} A = 13.02 \mu A

Explanation:

For this case we have the following info given:

Number of Na+ ions 5.7 x10^{11} ions

Each ion have a charge of +e and the crage of the electron is 1.6 x10^{-19}C

The time is given t = 7 ms if we convert this into seconds we got:

t = 7ms * \frac{1s}{1000 ms}= 0.007s

Now we can use the following formula given from the current passing thourhg a meter of nerve axon given by:

Q = Ne

Where N represent the number of ions, e the charge of the electron and Q the total charge

If we replace on this case we have this:

Q= 5.7x10^{11} * (1.6 x10^{-19}C) = 9.12x10^{-8} C

And from the general definition of current we know that:

I =\frac{Q}{t}

And since we know the total charge Q and the time we can replace:

I = \frac{9.12x10^{-8} C}{0.007 s}= 1.30285 x10^{-5} \frac{C}{s}=1.30285 x10^{-5} A = 13.02 \mu A

The current during the inflow charge in the meter axon for this case is 13.02 \mu A

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