Answer: The intensity level of sound in the bedroom is 80dB
Explanation:
Intensity of lawn mower at r=1m is 100dB
Beta1= 10dBlog(I1/Io)
100dB= 10dB log(I1/Io)
10^10= I1/Io
I1= Io(10^10)
10^12)×(10^10)= I1
I1=10^-2w/m^2
Intensity of lawn mower at r=20m
I2/I1=(r1/r2)^2 =(1/20)^2
I2= I1(1/400)
I2=2.5×10^-3W_m^2
Intensity of 4 lown mowers at 20m fro. Window
= 10dBlog(4I2/Io)
= 10^-4/10^-12
=80dB
For resistance we have R=ρ l/a
thus for conductance we have K=σ a/l
conductance,K=1/R
conductivity,σ =1/ρ
σ = .80 Ω-1 cm-1
l =9 cm
a = 3 cm²
K=.80 ×3/9
=0.26 Ω-1
Answer:
Bulk modulus = 1.35 ×
Pa
Explanation:
given data
density = 1400 kg/m³
frequency = 370 Hz
wavelength = 8.40 m
solution
we get here bulk modulus of the liquid that is
we know Bulk Modulus =
...............
here
is density i.e 1400 kg/m³
and v is = frequency × wavelength
v = 370 × 8.40 = 3108 m/s
so here bulk modulus will be as
Bulk modulus = 3108² × 1400
Bulk modulus = 1.35 ×
Pa
Change in momentum: finial momentum - initial momentum
Momentum = mass * velocity
Mass = 100g, same as 0.1kg
m(v-u) = 0.1(10-2) = 0.1(8)
The answer is 0.8Ns
Answer:
![V_3\approx 4.28\,\,V](https://tex.z-dn.net/?f=V_3%5Capprox%204.28%5C%2C%5C%2CV)
![I_1=0.0572\,\,amps](https://tex.z-dn.net/?f=I_1%3D0.0572%5C%2C%5C%2Camps)
![I_3\approx 0.171\,\,amps](https://tex.z-dn.net/?f=I_3%5Capprox%200.171%5C%2C%5C%2Camps)
Explanation:
Notice that this is a circuit with resistors R1 and R2 in parallel, connected to resistor R3 in series. It is what is called a parallel-series combination.
So we first find the equivalent resistance for the two resistors in parallel:
![\frac{1}{Re}= \frac{1}{R1}+\frac{1}{R2}\\\frac{1}{Re}= \frac{1}{100}+\frac{1}{50}\\\frac{1}{Re}= \frac{3}{100}\\Re=\frac{100}{3} \,\,\Omega](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BRe%7D%3D%20%5Cfrac%7B1%7D%7BR1%7D%2B%5Cfrac%7B1%7D%7BR2%7D%5C%5C%5Cfrac%7B1%7D%7BRe%7D%3D%20%5Cfrac%7B1%7D%7B100%7D%2B%5Cfrac%7B1%7D%7B50%7D%5C%5C%5Cfrac%7B1%7D%7BRe%7D%3D%20%5Cfrac%7B3%7D%7B100%7D%5C%5CRe%3D%5Cfrac%7B100%7D%7B3%7D%20%5C%2C%5C%2C%5COmega)
By knowing this, we can estimate the total current through the circuit,:
![Vs=I\,*\,(\frac{100}{3} +25)\\10=I\,*\,\frac{175}{3} \\I=\frac{30}{175} \,amps](https://tex.z-dn.net/?f=Vs%3DI%5C%2C%2A%5C%2C%28%5Cfrac%7B100%7D%7B3%7D%20%2B25%29%5C%5C10%3DI%5C%2C%2A%5C%2C%5Cfrac%7B175%7D%7B3%7D%20%5C%5CI%3D%5Cfrac%7B30%7D%7B175%7D%20%5C%2Camps)
So approximately 0.17 amps
and therefore, we can estimate the voltage drop (V3) in R3 uisng Ohm's law:
![V_3=\frac{30}{175} *\,25=\frac{30}{7} \approx 4.28\,\,V](https://tex.z-dn.net/?f=V_3%3D%5Cfrac%7B30%7D%7B175%7D%20%2A%5C%2C25%3D%5Cfrac%7B30%7D%7B7%7D%20%5Capprox%204.28%5C%2C%5C%2CV)
So now we know that the potential drop across the parellel resistors must be:
10 V - 4.28 V = 5.72 V
and with this info, we can calculate the current through R1 using Ohm's Law:
![I_1=\frac{V_1}{R_1} =\frac{5.72}{100} =0.0572\,\,amps](https://tex.z-dn.net/?f=I_1%3D%5Cfrac%7BV_1%7D%7BR_1%7D%20%3D%5Cfrac%7B5.72%7D%7B100%7D%20%3D0.0572%5C%2C%5C%2Camps)