Human ears can hear sound waves that vibrate in the range from about 20 times a second (a deep rumbling noise) to about 20,000 times a second (a high-pitched whistling). (Children can generally hear higher-pitched sounds than their parents, because our ability to hear high frequencies gets worse as we get older.) Speaking more scientifically, we could say that the sounds we can perceive have a frequency ranging from 20–20,000 hertz (Hz). A hertz is a measurement of how often something vibrates and 1 Hz is equal to one vibration each second. The human voice makes sounds ranging from a few hundred hertz to a few thousand hertz.Suppose you could somehow hit a drum-skin so often that it vibrated more than 20,000 times per second. You might be able to see the skin vibrating (just), but you certainly couldn't hear it. No matter how hard you hit the drum, you wouldn't hear a sound. The drum would still be transmitting sound waves, but your ears wouldn't be able to recognize them. Bats, dogs, dolphins, and moths might well hear them, however. Sounds this like, with frequencies beyond the range of human hearing, are examples of ultrasound.Infrasonics, vibrational or stress waves in elastic media, having a frequency below those of sound waves that can be detected by the human ear—i.e., below 20 hertz. The range of frequencies extends down to geologic vibrations that complete one cycle in 100 seconds or longer.
Answer:
1.991 × 10^(8) N/m²
Explanation:
We are told that its volume increases by 9.05%.
Thus; (ΔV/V_o) = 9.05% = 0.0905
To find the force per unit area which is also pressure, we will use bulk modulus formula;
B = Δp(V_o/ΔV)
Making Δp the subject gives;
Δp = B(ΔV/V_o)
Now, B is bulk modulus of water with a value of 2.2 × 10^(9) N/m²
Thus;
Δp = 2.2 × 10^(9)[0.0905]
Δp = 1.991 × 10^(8) N/m²
(a) The magnitude of the net force on the crate is -26.46 N.
(b) The net work done on the crate while it is on the rough surface is -17.2 J.
(c) The speed of the crate when it reaches the end of the rough surface is 0.61 m/s.
<h3>Magnitude of net force on the crate</h3>
The magnitude of the net force on the crate is calculated as follows;
F(net) = F - Ff
where;
F(net) = F - μmg
F(net) = 290 - (0.351)(92)(9.8)
F(net) = -26.46 N
<h3>Net work done on the crate</h3>
W = Fd
W = -26.46 x 0.65 = - 17.2 J
<h3>Speed of the crate </h3>
acceleration of the crate, a = F/m
a = -26.46/92
a = -0.288 m/s²
v² = u² + 2as
v² = 0.865² + (2)(-0.288)(0.65)
v² = 0.374
v = 0.61 m/s
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