600Hz is the driving frequency needed to create a standing wave with five equal segments.
To find the answer, we have to know about the fundamental frequency.
<h3>How to find the driving frequency?</h3>
- The following expression can be used to relate the fundamental frequency to the driving frequency;
f(n) = n * f (1)
where, f(1) denotes the fundamental frequency and the driving frequency f(n).
- The standing wave has four equal segments, hence with n=4 and f(n)=4, we may calculate the fundamental frequency.
f(4) = 4× f (1)
480 = 4× f(1)
f(1) = 480/4 =120Hz.
So, 120Hz is the fundamental frequency.
- To determine the driving frequency necessary to create a standing wave with five equally spaced peaks?
- For, n = 5,
f(n) = n 120Hz,
f(5) = 5×120Hz=600Hz.
Consequently, 600Hz is the driving frequency needed to create a standing wave with five equal segments.
Learn more about the fundamental frequency here:
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Ruby, a variety of the mineral corundum is in the trigonal crystal system, with hexagonal scalenohedra crystals
Answer:
it increases the amplitude of the wave as it propagates.
Explanation:
Answer:
502000W/m²
Heat flux = 800×50.2/20 =
502000W/m²
Explanation:
First, balance the reaction:
_ KClO₃ ==> _ KCl + _ O₂
As is, there are 3 O's on the left and 2 O's on the right, so there needs to be a 2:3 ratio of KClO₃ to O₂. Then there are 2 K's and 2 Cl's among the reactants, so we have a 1:1 ratio of KClO₃ to KCl :
2 KClO₃ ==> 2 KCl + 3 O₂
Since we start with a known quantity of O₂, let's divide each coefficient by 3.
2/3 KClO₃ ==> 2/3 KCl + O₂
Next, look up the molar masses of each element involved:
• K: 39.0983 g/mol
• Cl: 35.453 g/mol
• O: 15.999 g/mol
Convert 10 g of O₂ to moles:
(10 g) / (31.998 g/mol) ≈ 0.31252 mol
The balanced reaction shows that we need 2/3 mol KClO₃ for every mole of O₂. So to produce 10 g of O₂, we need
(2/3 (mol KClO₃)/(mol O₂)) × (0.31252 mol O₂) ≈ 0.20835 mol KClO₃
KClO₃ has a total molar mass of about 122.549 g/mol. Then the reaction requires a mass of
(0.20835 mol) × (122.549 g/mol) ≈ 25.532 g
of KClO₃.
