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:
In D: 3J
Explanation:
Potential energy: Ep=mgh where m is the mass, h altitude.
In point A: h=20cm=0.2m
Epa=12=0.2×mg. Thus mg=12/0.2=60N
For point D: hd=5cm=0.05m
Epd=mg×0.05=60×0.05=3J
It would probably be D. because that would be a waste of the water, and it's not helping conserve it.
If the input machine produces an
output of 80 Newtons, then the mechanical advantage is that it produces work. If
the required output must be 80 Newtons, then the input force is desirable
having 100 percent production of force even though it requires a large amount
of input to produce 80 Newtons of force.
Power = (voltage) x (current)
1 watt = (1 volt) x (1ampere)
= 1 "volt-amp" or 1 "amp-volt" .