Answer:
f_tympanum = 256 Hz.
Explanation:
The eardrum is the vibrating membrane of the human ear, it works by resonance, that is, an external stimulus (force) makes it vibrate, as the eardrum is extremely light it can vibrate at the same frequency of the incident sound.
Consequently if the incident vibration is f = 256 hz, the eardrum resonates at the same frequency
f_tympanum = 256 Hz.
As a reference the response of the eardrum goes from f = 20 Hz f = 20000 Hz
Answer:60 rev/min
Explanation:
Given
angular speed of first shaft 
Moment of inertia of second shaft is seven times times the rotational speed of the first i.e. If I is the moment of inertia of first wheel so moment of inertia of second is 7 I
As there is no external torque therefore angular momentum is conserved
This question is incomplete, the complete question is;
A parallel-plate capacitor is made from two aluminum-foil sheets, each 3.0 cm wide and 5.00 m long. Between the sheets is a mica strip of the same width and length that is 0.0225 mm thick. What is the maximum charge?
(The dielectric constant of mica is 5.4, and its dielectric strength is 1.00×10⁸ V/m)
Answer: the maximum charge q is 716.85 μF
Explanation:
Given data;
with = 3.0 cm = 0.03
breathe = 5.0 m
Area = 0.03 × 5 = 0.15 m²
dielectric strength E = 1.00 × 10⁸
∈₀ = 8.85 × 10⁻¹²
constant K = 5.4
maximum charge = ?
the capacitor C = KA∈₀ / d
q = cv so c = q/v
now
q/v = KA∈₀ / d
q = vKA∈₀/d = EKA∈₀
we substitute
q = (1.00 × 10⁸) × 5.4 × 0.15 × 8.85 × 10⁻¹²
q = 716.85 × 10⁻⁶ F
q = 716.85 μF
the maximum charge q is 716.85 μF
Answer:
44200 N
Explanation:
To calculate the average force exerted on the car, we will use the following equation
Where F is the average force, t is the time, m is the mass, vf is the final velocity and vi is the initial velocity of the car.
Replacing t = 0.5s, m = 1300 kg, vf = -2 m/s, and vi = 15 m/s and solving for F, we get
Therefore, the average force exerted on the car by the wall was 44200 N
Answer:
a) 0.162 V/m
b) 0.54 nT
c) 22 kW
Explanation:
Given
Distance of flight, d = 20 km
Intensity of signal, I = 35.1 μW/m²
Magnitude of the electric component is gotten using the formula
E(m) = √(2Iμc), where
E(m) = √(2 * 35*10^-6 * 4*3.142*10^-7 * 3*10^8)
E(m) = √(2 * 35*10^-6 * 1.257*10^-6 * 3*10^8)
E(m) = √0.0264
E(m) = 0.162 V/m
Magnitude of magnetic component can be gotten by using the relation
B(m) = E(m) / c
B(m) = 0.162 / 3*10^8
B(m) = 5.4*10^-10
B(m) = 0.54 nT
Transmission power, P = IA
where A = 1/2 * 4πr²
P = 2Iπr²
P = 2 * 35*10^-6 * 3.142 * (10000)²
P = 2 * 35*10^-6 * 3.142 * 1*10^8
P = 21994 W
Thus, the transmission power is 22 kW
