A) current
(I is always current in electricity)
Answer:
0.3 %
Explanation:
Earth cleans and replenishes the water supply through the hydrologic cycle. The earth has an abundance of water, but unfortunately, only a small percentage, is even usable by people.
Answer: Use this F=Ma.
Explanation: So your answer will be
F=1 Kg+9.8 ms-2
So the answer will be
F=9.8N
How'd I do this?
I just used Newton's second law of motion.
I'll also put the derivation just in case.
Applied force α (Not its alpha, proportionality symbol) change in momentum
Δp α p final- p initial
Δp α mv-mu (v=final velocity, u=initial velocity and p=v*m)
or then
F α m(v-u)/t
So, as we know v=final velocity & u= initial velocity and v-u/t =a.
So F α ma, we now remove the proportionality symbol so we'll add a proportionality constant to make the RHS & LHS equal.
So, F=<em>k</em>ma (where k is the proportionality constant)
<em>k</em> is 1 so you can ignore it.
So, our equation becomes F=ma
363 m/s is the speed of sound through the air in the pipe.
Answer: Option B
<u>Explanation:</u>
The formula used to calculate the wavelength given as below,
![Wavelength (\lambda)=\frac{\text { wave velocity }(v)}{\text { frequency }(f)}](https://tex.z-dn.net/?f=Wavelength%20%28%5Clambda%29%3D%5Cfrac%7B%5Ctext%20%7B%20wave%20velocity%20%7D%28v%29%7D%7B%5Ctext%20%7B%20frequency%20%7D%28f%29%7D)
--------> eq. 1
In power system, harmonics define by positive integers of the fundamental frequency. So the third order harmonic is a multiple of the third fundamental frequency. Each harmonic creates an additional node and an opposite node, as well as an additional half wave within the string.
If the number of waves in the circuit is known, the comparison between standing wavelength and circuit length can be calculated algebraically. The general expression for this given as,
![L=\frac{n \lambda}{2}](https://tex.z-dn.net/?f=L%3D%5Cfrac%7Bn%20%5Clambda%7D%7B2%7D)
For first harmonic, n =1
![L=\frac{\lambda}{2}](https://tex.z-dn.net/?f=L%3D%5Cfrac%7B%5Clambda%7D%7B2%7D)
For second harmonic, n =2
![L=\frac{2 \lambda}{2}=\lambda](https://tex.z-dn.net/?f=L%3D%5Cfrac%7B2%20%5Clambda%7D%7B2%7D%3D%5Clambda)
For third harmonic, n =3
![L=\frac{3 \lambda}{2}](https://tex.z-dn.net/?f=L%3D%5Cfrac%7B3%20%5Clambda%7D%7B2%7D)
-------> eq. 2
Here given f = 939 Hz, L = 0.58 m...And, substitute eq 2 in eq 1 and values, we get
![v=\frac{2 \times 0.58 \times 989}{3}=\frac{1089.24}{3}=363.08 \mathrm{m} / \mathrm{s}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B2%20%5Ctimes%200.58%20%5Ctimes%20989%7D%7B3%7D%3D%5Cfrac%7B1089.24%7D%7B3%7D%3D363.08%20%5Cmathrm%7Bm%7D%20%2F%20%5Cmathrm%7Bs%7D)
I think it’s the third one idk tho