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sweet [91]
3 years ago
15

Explain why the sound produced by every vibrating body cannot be heard by us ?​

Physics
1 answer:
VMariaS [17]3 years ago
5 0

Answer:

The human ear is not sensitive to every frequency of sound, rather, our hearing range is from 20Hz to 20,000Hz. This means sound frequencies outside this range are not audible to the human ear, which is why not every sound produced is heard. This is quite a blessing because if we could hear every vibrating body, then our ears would never stop listening: we won't be able to distinguish 'important sounds' from unimportant ones, rendering our ears effectively useless. Perhaps this is why we have a hearing range in the first place: sounds of our friends and our foes lie in 20Hz-20kHz range.

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Each driver has mass 79.0 kg. Including the masses of the drivers, the total masses of the vehicles are 800 kg for the car and 4
Mademuasel [1]

Answer:

Force exerted on the car driver by the seatbelt = 8139.4 N = 8.14 kN

Force exerted on the truck driver by the seatbelt = 1628.2 N = 1.63 kN

It is evident that the driver of the smaller vehicle has it worse. The car driver is in way more danger in this perfectly inelastic head-on collision with a bigger vehicle (the truck).

Explanation:

First of, we calculate the velocity of the vehicles after collision using the law of conservation of Momentum

Momentum before collision = Momentum after collision

Since the collision of the two vehicles was described as a head-on collision, for the sake of consistent convention, we will take the direction of the velocity of the bigger vehicle (the truck) as the positive direction and the direction of the car's velocity automatically is the negative direction.

Velocity of the truck before collision = 6.80 m/s

Velocity of the car before collision = -6.80 m/s

Let the velocity of the inelastic unit of vehicles after collision be v

Momentum before collision = (4000)(6.80) + (800)(-6.80) = 27200 - 5440 = 21,760 kgm/s

Momentum after collision = (4000 + 800)(v) = (4800v) kgm/s

Momentum before collision = Momentum after collision

21760 = 4800v

v = (21760/4800)

v = 4.533 m/s (in the direction of the big vehicle (the truck)

So, we then apply Newton's second law of motion which explains that the magnitude change in momentum is equal to the magnitude of impulse.

|Impulse| = |Change in momentum|

But Impulse = (Force exerted on each driver by the seatbelt) × (collision time) = (F×t)

Change in momentum = (Momentum after collision) - (Momentum before collision)

So, for the driver of the truck

Initial velocity = 6.80 m/s (the driver moves with the velocity of the truck)

Final velocity = 4.533 m/s

Change in momentum of the truck driver = (79)(6.80) - (79)(4.533) = 179.1 kgm/s

(F×t) = 179.1

F × 0.110 = 179.1

F = (179.1/0.11)

F = 1628.2 N = 1.63 kN

So, for the driver of the car

Initial velocity = -6.80 m/s (the driver moves with the velocity of the car)

Final velocity = 4.533 m/s

Change in momentum of the car driver = (79)(-6.80) - (79)(4.533) = -895.3 kgm/s

(F×t) = |-895.3|

F × 0.110 = 895.3

F = (895.3/0.11)

F = 8139.4 N = 8.14 kN

Hope this Helps!!!

3 0
3 years ago
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Answer:

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Explanation:

The graph that curves downwards is negative acceleration. While the position decreases the slop increases.

8 0
2 years ago
What makes west region different?
anzhelika [568]

Answer:

The West is known for "wide, open spaces", cattle, mines, and mountains.

7 0
3 years ago
There is much volcanic activity in Iceland, and the country straddles the Mid-Atlantic Ridge. Therefore, Iceland is situated ___
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3 years ago
At some distance from a point charge, the electric potential is 635.0 V and the magnitude of the electric field is 189.0 N/C. Fi
Nataliya [291]

Answer:

The distance from the charge is 3.35 m.                

Explanation:

Given that,

Electric potential, V = 635 V

Magnitude of electric field, E = 189 N/C

We need to find the distance from the charge. We know that the relation between electric field and electric potential is given by :

E=\dfrac{V}{d}

d is the distance from charge

d=\dfrac{V}{E}\\\\d=\dfrac{635}{189}\\\\d=3.35\ m

So, the distance from the charge is 3.35 m. Hence, this is the required solution.

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