<span>Ocean currents act much like a conveyer belt,
transporting warm water and precipitation from the equator toward the
poles and cold water from the poles back to the tropics. Thus, currents
regulate global climate, helping to counteract the uneven distribution of solar radiation reaching Earth's surface.</span>
Answer:
- quality factor (Q) = 69.99
- inductor = 1.591 x 10⁻⁴ H
- capacitor = 3.248 x 10⁻¹⁰ F
Explanation:
Given;
resonance frequency (F₀) = 700 kHz
resistor, R = 10 Ohm
bandwidth (BW) = 10 kHz
bandwidth (BW) 

make L (inductor) the subject of the formula


make C (capacitor) the subject of the formula

quality factor (Q) 
quality factor (Q) = 69.99
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:
c) F = 16000 N
Explanation:
For this exercise we use Newton's second law
F = ma
they tell us that adding the other wagons the acceleration of the locomotive must be maintained
F = m a
by adding the other four wagons
mass = 4 no
therefore to maintain the force you must also raise the same factor
Fe = 4Fo
Fe = 4 4000
F = 16000 N