912.
outer ear:
pinna
ear canal
middle ear:
ossicles and ear drum
inner ear:
semcircular canals
cochlea
auditory nerve
13.
frequency = wavespeed ÷ wavelength
14.
if frequency increases you would experience a higher pitch in sound
15.
humans can hear 20Hz to 20kHz
16.
The Doppler effect is the change in frequency or wavelength of a wave for an observer who is moving relative to the wave source. Can be used for machines measuring speed via doppler effect.
17.
Doppler in hospitals can be used for ultrasound to provide images for diagnosis and monitoring.
Answer:
w = vR/3
Explanation:
The centre of mass of the loop to bullet system is given by D / 4 from centre of loop, which is equivalent to R / 2 from its centre.
From the principle of conservation of linear momentum
, we have
m*v = 2*m* Vcm
Where v = velocity of bullet, Vcm = velocity of wood
Hence, we have
Vcm = v2
Also, from the conservation of angular momentum about the centre of mass.
M*V*(R/2) = Ic*w - equation (I)
where Ic = moment of inertia and w = angular velocity
Ic for a ring is given by 
Ic of a bullet is given by 
Hence, the moment of inertia of the system is given by the summation of the two moments of inertia Ic(ring) + Ic(bullet) which gives
Ic(system) = 
Substituting back into equation (I), we have
Hence, we obtain w =vR/3
w=v3R
<span>B. velocity .................</span>
Answer:
a) 0.167 μC/m^2
b) 1.887 * 10^4 V/m
Explanation:
Hello!
First let's find the surface charge density:
a)
Since thesatellite is metallic, the accumalted charge will be uniformly distribuited on its surface. Therefore the charge density σ will be:
σ = Q/A
Where A is the area of the satellite, which is:
A=4πr^2 = πd^2 = π(1.9m)^2
Therefore:
σ = (1.9)/(π (1.9)^2) μC/m^2 = 0.167 μC/m^2
Now let's calculate the electric field
b)
Just outside the surface of the satellite the elctric field will be:
E = σ/ε0
Where ε0=8.85×10^−12 C/Vm
Therefore:
E = (0.167*10^-6 C/m^2) / (8.85*10^-12 C/Vm) = 0.01887 *10^6 V/m
E = 1.887 * 10^4 V/m
Answer:
A : hot and moist, maritime tropical
B: cold and dry, maritime polar
C: hot and moist , maritime tropical
D: cold and dry, continental polar
E: hot and moist , maritime tropical
F: cold and dry , maritime polar
Explanation:
Cold air is denser than warm air. The more water vapor that is in the air, the less dense the air becomes. That is why cold, dry air is much heavier than warm, humid air.
Maritime polar (mP) air masses are cool, moist, and unstable. Some maritime polar air masses originate as continental polar air masses over Asia and move westward over the Pacific, collecting warmth and moisture from the ocean.
Maritime tropical (mT) air masses are warm, moist, and usually unstable.
