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sweet-ann [11.9K]
3 years ago
5

The function of the ossicles in the middle ear is to transmit the vibrations of the tympanic membrane caused by sound waves prop

agated in air to the fluid-filled cochlea. In doing so, the ossicles change the nature of the vibrations. The ear tries to conserve the energy in the wave I,
where v is the wave speed, cramster-equation-2010731130406341613664 is the density of the medium, cramster-equation-2010731131286341613668 is the frequency of the wave and A is the amplitude of the wave.
Calculate the magnitude of the change in A that occurs. The density of air and water is 1.3x10–3g/cm3 and 1.0g/cm3, respectively, and the velocity of sound is 331 m/s in air and 1410 m/s in water. The frequency of the wave remains constant as the wave propagates. (Hint: Begin with to find .)
Physics
1 answer:
Daniel [21]3 years ago
6 0
A I hope I helpedddd
You might be interested in
1.2
Alika [10]

Answer:

B

Explanation:

Because it has to increase

3 0
3 years ago
Read 2 more answers
A charge +q is located at the origin, while an identical charge is located on the x axis at x = +0.57 m. A third charge of +2q i
Jlenok [28]

Answer:

The third charge placed is 0.80 m.

Explanation:

Given that,

Distance = 0.57 m

First charge = q

Third charge = 2q

We need to calculate the electrostatic force on charge q₁ due to q₂

Using formula of electrostatic force

F_{21}=\dfrac{kq_{1}q_{2}}{r_{1}^2}

When placed another charge q₃ at certain distance from origin, then the net force on charge q₁ due to both charges is

F_{net}=F_{21}+F_{31}

The net electrostatic force on the charge at the origin doubles.

2F_{21}=F_{21}+F_{31}

F_{31}=F_{21}

\dfrac{kq_{3}q_{1}}{r_{2}^2}=\dfrac{kq_{2}q_{1}}{r_{1}^2}

\dfrac{q_{3}}{r_{2}^2}=\dfrac{q_{2}}{r_{1}^2}

r_{2}^2=\dfrac{q_{3}}{q_{2}\timesr_{1}^2}

r_{2}=\sqrt{\dfrac{q_{3}}{q_{2}}}r_{1}

Put the value into the formula

r_{2}=\sqrt{\dfrac{2q}{q}}\times0.57

r_{2}=\sqrt{2}\times0.57

r_{2}=0.80\ m

Hence, The third charge placed is 0.80 m from origin in x-axis.

4 0
3 years ago
A negative slope on the velocity versus time graph indicates that an object is not accelerating true or false ?
dangina [55]

Explanation :

We know that the slope of velocity -time graph gives the acceleration. Acceleration of an object is defined as the rate of change of velocity i.e.

a=\dfrac{dv}{dt}

Suppose a driver suddenly applies brakes. In this case the initial velocity of his or her vehicle is more and the final velocity is less.

So, the acceleration is negative in this case i.e. the object is decelerating.

A negative slope on the velocity versus time graph indicates that an object is not accelerating. This statement is false as the object is decelerating.

8 0
3 years ago
A lump of steel of mass 10kg at 627 degree Celsius is dropped in 100kg oil at 30 degree Celsius . the specific heat of steel And
Naily [24]

Answer:

700J

Explanation:

8 0
3 years ago
car 1 is traveling south at 18 m/s and has a full load, giving it a total mass of 14,650 kg. Car 2 is traveling north at 11 m/s
kvasek [131]

Answer:

v_{4}= 80.92[m/s] (Heading south)

Explanation:

In order to calculate this problem, we must use the linear moment conservation principle, which tells us that the linear moment is conserved before and after the collision. In this way, we can propose an equation for the solution of the unknown.

ΣPbefore = ΣPafter

where:

P = linear momentum [kg*m/s]

Let's take the southward movement as negative and the northward movement as positive.

-(m_{1}*v_{1})+(m_{2}*v_{2})=-(m_{1}*v_{3})+(m_{2}*v_{4})

where:

m₁ = mass of car 1 = 14650 [kg]

v₁ = velocity of car 1 = 18 [m/s]

m₂ = mass of car 2 = 3825 [kg]

v₂ = velocity of car 2 = 11 [m/s]

v₃ = velocity of car 1 after the collison = 6 [m/s]

v₄ = velocity of car 2 after the collision [m/s]

-(14650*18)+(3825*11)=(14650*6)-(3825*v_{4})\\v_{4}=80.92[m/s]

4 0
3 years ago
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