Answer:
5 ms-2
Explanation:
F = ma
F = 100N
m = 20kg ( you should make sure the unit is kg before you answer the question)
100 = 20a
a = 100÷ 20
a = 5 ms-2
Angry sound level = 70 db
Soothing sound level = 50 db
Frequency, f = 500 Hz
Assuming speed of sound = 345 m/s
Density (assumed) = 1.21 kg/m^3
Reference sound intensity, Io = 1*10^-12 w/m^2
Part (a): Initial sound intensity (angry sound)
10log (I/Io) = Sound level
Therefore,
For Ia = 70 db
Ia/(1*10^-12) = 10^(70/10)
Ia = 10^(70/10)*10^-12 = 1*10^-5 W/m^2
Part (b): Final sound intensity (soothing sound)
Is = 50 db
Therefore,
Is = 10^(50/10)*10^-12 = 18*10^-7 W/m^2
Part (c): Initial sound wave amplitude
Now,
I (W/m^2) = 0.5*A^2*density*velocity*4*π^2*frequency^2
Making A the subject;
A = Sqrt [I/(0.5*density*velocity*4π^2*frequency^2)]
Substituting;
A_initial = Sqrt [(1*10^-5)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-8 m = 69.7 nm
Part (d): Final sound wave amplitude
A_final = Sqrt [(1*10^-7)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-9 m = 6.97 nm
Answer:
8.5 Ω
Explanation:
La resistencia de un material es directamente proporcional a su longitud e inversamente proporcional al área de la sección transversal.
La fórmula de la resistencia (R) viene dada por:
R = ρL/A
Donde ρ es la resistividad del material, L es la longitud del material y A es el área de la sección transversal del material.
Dado que:
L = 1 km = 1000 m, A = 2 mm² = 2 * 10⁻⁶ m², ρ (cobre) = 1.7 * 10⁻⁸ Ωm
Sustituyendo da:
R = 1,7 * 10⁻⁸ * 1000/2 * 10⁻⁶
R = 8.5 Ω
Explanation:
It is given that,
Bandwidth of a laser source, 
(b) Let t is the time separation of sections of sections of the light wave that can still interfere. The time period is given by :
(a) Let h is the coherence length of the source. It is given by :
c is the speed of light
l = 0.0099 m
Hence, this is the required solution.
Answer:
0.125 m
Explanation:
In this problem, we have:
v = 0.50 m/s is the average velocity of the wave
T = 0.25 s is the period of the wave
We can find the frequency of the wave, which is equal to the reciprocal of the period:
The problem is asking us to find the distance between two crests of the wave: this is equivalent to the wavelength. The wavelength is related to the average velocity and the frequency by the formula:
Substituting the numerical values, we find
