The specific heat of mercury is 149.4 J/(kgK)
Explanation:
When a substance is supplied with an amount of energy Q, its temperature increases according to the equation:
where
is the increase in temperature
m is the mass of the sample
is its specific heat capacity
For the sample of mercury in this problem we have
Q = 275 J
m = 0.450 kg
Therefore, by re-arranging the equation we find the mercury's specific heat:
Learn more about specific heat capacity:
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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.
Answer:
Explanation:
From the question we are told that:
Acceleration
Displacement
Initial time
Final Time
Generally the equation for Velocity of 1.05 travel is mathematically given by
Using Newton's Law of Motion
Generally the equation for Distance traveled before stop is mathematically given by
Generally the equation for Distance to stop is mathematically given by
Since For this Final section
Final velocity
Initial velocity
Therefore
Using Newton's Law of Motion
Giving
Therefore
Generally the Total Distance Traveled is mathematically given by