Answer: I organized the data collected from the lab by using a table for each test. Table A contains the data collected from the table tennis ball and the golf ball, Table B contains the data collected from the inflated football and the deflated football, and Table C contains data collected from the question baseball and the data from Table C helps to find out whether or not the baseball is legitimate or altered.
Explanation:
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<span>f(x) = 5.05*sin(x*pi/12) + 5.15
First, you need to determine the period of the function. The period will be the time interval between identical points on the sinusoidal function. For this problem, the tide is rising and at 5.15 at midnight for two consecutive days. So the period is 24 hours. Over that 24 hour period, we want the parameter passed to sine to range from 0 to 2*pi. So the scale factor for x will be 2*pi/24 = pi/12 which is approximately 0.261799388. The next thing to note is the magnitude of the wave. That will simply be the difference between the maximum and minimum values. So 10.2 ft - 0.1 ft = 10.1 ft. And since the value of sine ranges from -1 to 1, we need to divide that magnitude by 2, so 10.1 ft / 2 = 5.05 ft.
So our function at this point looks like
f(x) = 5.05*sin(x*pi/12)
But the above function ranges in value from -5.05 to 5.05. So we need to add a bias to it in order to make the low value equal to 0.1. So 0.1 = X - 5.05, 0.1 + 5.05 = X, 5.15 = X. So our function now looks like:
f(x) = 5.05*sin(x*pi/12) + 5.15
The final thing that might have been needed would have been a phase correction. With this problem, we don't need a phase correction since at X = 0 (midnight), the value of X*pi/12 = 0, and the sine of 0 is 0, so the value of the equation is 5.15 which matches the given value of 5.15. But if the problem had been slightly different and the height of the tide at midnight has been something like 7 feet, then we would have had to calculate a phase shift value for the function and add that constant to the parameter being passed into sine, making the function look like:
f(x) = 5.05*sin(x*pi/12 + C) + 5.15
where
C = Phase correction offset.
But we don't need it for this problem, so the answer is:
f(x) = 5.05*sin(x*pi/12) + 5.15
Note: The above solution assumes that angles are being measured in radians. If you're using degrees, then instead of multiplying x by 2*pi/24 = pi/12, you need to multiply by 360/24 = 15 instead, giving f(x) = 5.05*sin(x*15) + 5.15</span>
Wind speed and air temperature are used to calculate a windchill factor.
<u>Explanation:</u>
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Wind-chill factor is the reduction of body temperature due to the passing flow of lower-temperature air.
The air temperature value is always higher than the wind chill numbers. the heat index will be used if the apparent temperature is higher than the air temperature.So, Wind speed and air temperature are mainly used to calculate a windchill factor.
There are many ways, the surface loses its heat through conduction, evaporation,radiation, and convection.The rate of convection depends on the difference in temperature between the surface and the fluid surrounding the surface and the velocity of that fluid with respect to the surface. The air around the warm surface will be heated, an insulating layer of warm air forms against the surface.The layer becomes a boundary between two. As the wind speed is high the surface cools down rapidly.
A- reflection B- refraction C- refraction D- reflection E- reflection F- refraction