No, aluminum has a density near 2.7 g/cm^3
<span>7.8 g/cm^3 is near the density of iron (or in the case of a fork, steel).
this is it
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From what we know, we can confirm that this ratio (turning up the volume by one click relative to the TV's overall volume) can be quantified as the Weber fraction.
<h3>What is the Weber fraction?</h3>
This fraction describes the ratio needed for change to a stimulus in which the change is just barely noticeable. This question is a prime example in that it seeks to find out just how low of a difference is needed in TV volume in order for the difference to be noticeable.
Therefore, we can confirm that this ratio (turning up the volume by one click relative to the TV's overall volume) can be quantified as the Weber fraction.
To learn more about Weber visit:
brainly.com/question/5004433?referrer=searchResults
Answer:
0.546 
Explanation:
From the given information:
The force on a given current-carrying conductor is:

where the length usually in negative (x) direction can be computed as

Now, taking the integral of the force between x = 1.0 m and x = 3.0 m to get the value of the force, we have:



![F = I (9.0) \bigg [\dfrac{x^3}{3} \bigg ] ^3_1 \hat k](https://tex.z-dn.net/?f=F%20%3D%20I%20%20%289.0%29%20%5Cbigg%20%5B%5Cdfrac%7Bx%5E3%7D%7B3%7D%20%5Cbigg%20%5D%20%5E3_1%20%5Chat%20k)
![F = I (9.0) \bigg [\dfrac{3^3}{3} - \dfrac{1^3}{3} \bigg ] \hat k](https://tex.z-dn.net/?f=F%20%3D%20I%20%20%289.0%29%20%5Cbigg%20%5B%5Cdfrac%7B3%5E3%7D%7B3%7D%20-%20%5Cdfrac%7B1%5E3%7D%7B3%7D%20%5Cbigg%20%5D%20%20%5Chat%20k)
where;
current I = 7.0 A
![F = (7.0 \ A) (9.0) \bigg [\dfrac{27}{3} - \dfrac{1}{3} \bigg ] \hat k](https://tex.z-dn.net/?f=F%20%3D%20%287.0%20%5C%20A%29%20%20%289.0%29%20%5Cbigg%20%5B%5Cdfrac%7B27%7D%7B3%7D%20-%20%5Cdfrac%7B1%7D%7B3%7D%20%5Cbigg%20%5D%20%20%5Chat%20k)
![F = (7.0 \ A) (9.0) \bigg [\dfrac{26}{3} \bigg ] \hat k](https://tex.z-dn.net/?f=F%20%3D%20%287.0%20%5C%20A%29%20%20%289.0%29%20%5Cbigg%20%5B%5Cdfrac%7B26%7D%7B3%7D%20%5Cbigg%20%5D%20%20%5Chat%20k)
F = 546 × 10⁻³ T/mT 
F = 0.546 