<span>v is perpendicular to both E and B and has a magnitude E/B</span>
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:
freqiency=velocity/wavelength
ief=20/2000=0.01hz
Let the time be t.
so,In time t , distance travelled by 1st cyclist = 12.6 t
distance travelled by 2nd cyclist = 9.2t + 0.5 (1.5) t^2
Now, cyclist 1st is already 11.4 m ahead of 2nd cyclist.
so, 9.2t + 0.5 (1.5) t^2 = 11.4 + 12.6t
find t :
t = 6.77 sec
Answer:
Explanation:
For this problem we must use Newton's second law where force is gravitational attraction
F = m a
Since movement is circular, acceleration is centripetal.
a = v2 / r
Let's replace
G m M / r² = m v² / r
G M r = v²
The distance r is from the center of the planet
r = R + h
v = √ GM / (R + h)
If the friction force is not negligible
F - fr = m a
Where the friction force must have some functional relationship, for example
Fr = b v + c v² +…
Suppose we are high enough for the linear term to derive the force of friction
G m M / r - (m b v + m c v2) = m v2
G M / r - b v = v²
We see that the solution of the problem gives lower speeds and that change over time.
To compensate for this friction force, the motors should be intermittently suspended to recover speed.
