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jok3333 [9.3K]
2 years ago
12

Which structure is part of the inner ear?

Physics
2 answers:
ValentinkaMS [17]2 years ago
8 0

cochlea is part of the inner ear. have a great day my friend

Ulleksa [173]2 years ago
3 0

Answer:

Which structure is part of the inner ear?  

anvil  

eardrum  

ear canal  

<u>cochlea</u>

Hope this helps :)

Have a great day

5INGH

Explanation:

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Apparent brightness depends on which two properties?i’ll give brainlist?:)
Norma-Jean [14]

Answer:

D

Explanation:

I just did it on a p e x

8 0
2 years ago
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A waterbed has a force of 1300N on the floor. It exerts 347 Pa of pressure. What is the area of the waterbed?
Sergio [31]

Answer:

Pressure = Force/Area

347 = 1300/Area

Area = 1300/347

area= 3.47m2

6 0
3 years ago
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At what point in its path does a baseball that is hit to the outfield have its minimum vertical speed?
ioda

Answer:

At the point when it reaches its maximum height (answer C)

Explanation:

As soon as the ball leaves the bat is has not only horizontal component of the velocity, but also vertical, which allows it travel upwards in its trajectory.

When it reaches its maximum elevation, and the trajectory starts curving down, is when the velocity of the baseball is just horizontal (the vertical component has been reduced to zero due to the constant action of the acceleration of gravity which was slowing down it ascending pattern.

After that point, the ball starts gaining now vertical velocity being accelerated towards the center of the Earth due to gravity. At the maximum height of its path is when the vertical component of the velocity switches from positive (pointing up) to negative (pointing down) going through zero magnitude.

5 0
3 years ago
Which of these is always true about a real image?
hammer [34]
B. The image is always enlarged
6 0
3 years ago
Elevated water tanks are used in a municipal water distribution system to provide adequate pressure. Compute the height (h) of t
Lemur [1.5K]

Answer:

a)  h = 53.8 m,  b)   h_minimum = 28 m, h_maximum = 63.3 m

Explanation:

a) For this exercise let's use Bernoulli's equation.

The subscript 1 is for the tank and the subscript for the building

          P₁ + ½ ρ g v₁² + ρ g y₁ = P₂ + ½ ρ g v₂² + ρ g y₂

In general, the water tanks are open to the atmosphere, so P1 = Patm, also the tanks are very large so the speed of the water surface is very small v₁=0 and as they give us the precious static, this it is when the keys are closed so the output velocity is zero, v₂= 0. The height of the floors in a building is y₂ = 12 m

           

we substitute in Bernoulli's equation

         P_{atm} + 0 + ρ g h = P₂ + 0 + ρ g y₂

         h = \frac{(P_2 - P_{atm}) + \rho \ g \ y_2}{\rho \ g}

         h = \frac{\Delta P}{\rho g} + y₂

indicate that the value of ΔP = 410 10³ Pa

       

we calculate

           h = 410 10³ / (1000 9.8) + 12

           h = 53.8 m

b) ask for the height range for the minimum and maximum pressure

            h =\frac{\Delta P}{\rho g} ΔP / rho g

minimum

           h_minimum = 275 103/1000 9.8

           h_minimum = 28 m

maximums

           h_maximo = 620 103/1000 9.8

           h_maximum = 63.3 m

8 0
3 years ago
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