-Reduce the sample size so the experiment can be done faster.-Increase the sample size from 6 cups to 12 cups of sand and water.-Use more legible handwriting when recording data.-Use more precise digital thermometers.<span>-Use more precise scales that measure to the hundredth of a gram.</span>
Answer:
W = 3.12 J
Explanation:
Given the volume is 1.50*10^-3 m^3 and the coefficient of volume for aluminum is β = 69*10^-6 (°C)^-1. The temperature rises from 22°C to 320°C. The difference in temperature is 320 - 22 = 298°C, so ΔT = 298°C. To reiterate our known values we have:
β = 69*10^-6 (°C)^-1 V = 1.50*10^-3 m^3 ΔT = 298°C
So we can plug into the thermal expansion equation to find ΔV which is how much the volume expanded (I'll use d instead of Δ because of format):
So ΔV = 3.0843*10^-5 m^3
Now we have ΔV, next we have to solve for the work done by thermal expansion. The air pressure is 1.01 * 10^5 Pa
To get work, multiply the air pressure and the volume change.
W = 3.12 J
Hope this helps!
Answer:
Electrolysis is the process by which ionic substances are decomposed (broken down) into simpler substances when an electric current is passed through them. ... Electricity is the flow of electrons or ions. For electrolysis to work, the compound must contain ions.
Answer:
Some of the frequency that cannot be produced by the string includes 400Hz, 500Hz 650Hz etc...
Explanation:
Harmonics in strings are defined as the integral multiples of its fundamental frequency. This multiples are in arithmetic progression.
For example if Fo is the fundamental frequency of the string, the harmonics will be 2fo, 3fo, 4fo, 5fo... etc
If the string produces a fundamental frequency of 150Hz, some of the harmonics produced by the string will be 300Hz, 450Hz, 600Hz, 750Hz... etc
Some of the harmonics that cannot be produced include 400Hz, 500Hz 650Hz etc...
We are given with the x and y components of Vector A and B. In this case, we compute the resultant of both components of each vector. The vector is equal to the square root of the sum of the squares of the components. A is equal to 2.60 cm. B is equal to 5.56 cm. B is found in quadrant Iv and has an angle of 42.447 degrees as a terminal angle. A has an angle of 59.98 degrees.
a. 5.6082 < -15.53 degreesc. 6.63 <-64.98 degreesb. x = 6.63 cos -64.98 degrees = 2.80 y = 6.63 sin -64.98 degrees = -6.00
