The net force on a 0.50-cm-diameter eardrum is mathematically given as
F= 0.76969 N
<h3>What is the net force on a 0.50-cm-diameter eardrum?</h3>
Generally, the equation for Pressure is mathematically given as
P = ρgh
Therefore
P= 1000*9.8*4
P= 39200 Pa
Where
A= pi*(0.005/2)^2
Generally, the equation for Net force is mathematically given as
F = PA
F= 39200 *( pi*(0.005/2)^2)
F= 0.76969 N
In conclusion, The net force is
F= 0.76969 N
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Answer:
LOL
Explanation:
IMAGINE POSTING UR CLASSWORK LOLL
Answer:
See the answers below.
Explanation:
The cost of energy can be calculated by multiplying each given value, a dimensional analysis must be taken into account in order to calculate the total value of the cost in Rs.
![Cost=0.350[kW]*12[\frac{hr}{1day}]*30[days]*4.5[\frac{Rs}{kW*hr} ]=567[Rs]](https://tex.z-dn.net/?f=Cost%3D0.350%5BkW%5D%2A12%5B%5Cfrac%7Bhr%7D%7B1day%7D%5D%2A30%5Bdays%5D%2A4.5%5B%5Cfrac%7BRs%7D%7BkW%2Ahr%7D%20%5D%3D567%5BRs%5D)
The fuse can be calculated by knowing the amperage.
![P=V*I](https://tex.z-dn.net/?f=P%3DV%2AI)
where:
P = power = 350 [W]
V = voltage = 240 [V]
I = amperage [amp]
Now clearing I from the equation above:
![I=P/V\\I=350/240\\I=1.458[amp]](https://tex.z-dn.net/?f=I%3DP%2FV%5C%5CI%3D350%2F240%5C%5CI%3D1.458%5Bamp%5D)
The fuse should be larger than the current of the circuit, i.e. about 2 [amp]
Force is mass times acceleration. This means an object with a larger mass needs a stronger force to be moved along at the same acceleration as an object with a small mass
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)