A gas is contained in a vertical piston-cylinder assembly by a piston with a face area of 40 in2 and weight of 100 lbf. The atmo
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Answer:
The rule for kilometers is that every three seconds between a lightning flash and the following thunder gives the distance to the flash in kilometers.
Explanation:
In order to use the rule of thumb to find the speed of sound in meters per second, we need to use some conversion ratios. We know there is 1 mile per every 5 seconds after the lightning is seen. We also know that there are 5280ft in 1 mile and we also know that there are 0.3048m in 1ft. This is enough information to solve this problem. We set our conversion ratios like this:
notice how the ratios were written in such a way that the units got cancelled when calculating them. Notice that in one ratio the miles were on the numerator of the fraction while on the other they were on the denominator, which allows us to cancel them. The same happened with the feet.
The problem asks us to express the answer to one significant figure so the speed of sound rounds to 300m/s.
For the second part of the problem we need to use conversions again. This time we will write our ratios backwards and take into account that there are 1000m to 1 km, so we get:
This means that for every 3.11s there will be a distance of 1km from the place where the lightning stroke. Since this is a rule of thumb, we round to the nearest integer for the calculations to be made easily, so the rule goes like this:
The rule for kilometers is that every three seconds between a lightning flash and the following thunder gives the distance to the flash in kilometers.
Starting from the angular velocity, we can calculate the tangential velocity of the stone:
Then we can calculate the angular momentum of the stone about the center of the circle, given by
where
m is the stone mass
v its tangential velocity
r is the radius of the circle, that corresponds to the length of the string.
Substituting the data of the problem, we find
Then we can calculate the angular momentum of the stone about the center of the circle, given by
where
m is the stone mass
v its tangential velocity
r is the radius of the circle, that corresponds to the length of the string.
Substituting the data of the problem, we find