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3 years ago

Describe how the human ear hears sound.

Physics

1 answer:

fiasKO [112]3 years ago

8 0

Answer:

sound waves enter the outer ear and travel through a narrow passageway called the ear canal, which leads to the eardrum. The eardrum vibrates from the incoming sound waves and sends these vibrations to three tiny bones in the middle ear.

Explanation:

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List of priceless your bodies from largest to smallest in terms of their distance from earth

Marina86 [1]

Answer:

Explanation:

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3 years ago

Kyle lays a mirror flat on the floor and aims a laser at the mirror. The laser beam reflects off the mirror and strikes an adjac

Elena L [17]

Answer:

The angle of incidence is $\theta_i =42.6^o$

Explanation:

The diagram for this question is shown on the first uploaded image

From the question we are told that

The distance between the mirror and the wall is $a = 33.7 cm$

The height of the above the mirror is $b = 36.7 cm$

Generally the angle which the reflected ray make with the mirror is mathematically evaluated as

$\alpha =tan ^{-1} (\frac{b}{a})$

substituting values

$\alpha = tan ^{-1}( \frac{36.7}{33.7})$

$\alpha =47.4^o$

From the diagram we can deduce that the angle of incidence is

$\theta_i = 90 - \alpha$

So $\theta_i = 90 - 47.4$

$\theta_i =42.6^o$

5 0

3 years ago

A boy throws a ball vertically up it returns the ground after 10 seconds find the maximum height reached by the ball

Akimi4 [234]

Answer:

Approximately $122.625\; {\rm m}$ (assuming that $g = 9.81\; {\rm m\cdot s^{-2}}$ , the ball was launched from ground level, and that the drag on the ball is negligible.)

Explanation:

Let $v_{0}$ denote the velocity at which the ball was thrown upward.

If the drag (air friction) on the ball is negligible, the ball would land with a velocity of exactly $(-v_{0})$ . The velocity of the ball would be changed from $v$ to $(-v_{0})\!$ (such that $\Delta v = (-v_{0}) - v_{0} = (-2\, v_{0})$ ) within $t = 10\; {\rm s}$ .

Also because the drag on the ball is negligible, the acceleration of the ball would be $a = -g = -9.81\; {\rm m\cdot s^{-2}}$ . Thus:

$\Delta v = a\, t = 10\; {\rm s} \times (-9.81\; {\rm m\cdot s^{-2}}) = -98.1\; {\rm m\cdot s^{-1}}$ .

Since $\Delta v = (-2\, v_{0})$ :

$-2\, v_{0} = \Delta v = -98.1\; {\rm m\cdot s^{-1}$ .

$\begin{aligned}v_{0} &= \frac{-98.1\; {\rm m\cdot s^{-1}}}{-2}= 49.05\; {\rm m \cdot s^{-1}}\end{aligned}$ .

The ball reaches maximum height when its velocity is $v_{1} = 0\; {\rm m\cdot s^{-1}}$ . Apply the SUVAT equation $x = ({v_{1}}^{2} - {v_{0}}^{2}) / (2\, a)$ to find the displacement $x$ between the original position (ground level, where $v_{0} = 49.05\; {\rm m\cdot s^{-1}}$ ) and the max-height position of the ball (where $v_{1} = 0\; {\rm m\cdot s^{-1}}$ .)

$\begin{aligned}x &= \frac{(0\; {\rm m\cdot s^{-1}})^{2} - (49.05\; {\rm m\cdot s^{-1}})^{2}}{2 \times (-9.81\; {\rm m\cdot s^{-2}})} \\ &\approx 122.625\; {\rm m\cdot s^{-1}}\end{aligned}$ .

7 0

2 years ago

An electric lamp creates light by forcing electrons through the filament of its light bulb. The kinetic energy of the moving ele

ollegr [7]

electric potential energy..electrons get the the kinetic energy from the voltage applied across the conductor

6 0

3 years ago

Read 2 more answers

A particle moving along the x-axis has its position described by the function x=(3.00t3−1.00t 2.00)m where t is in s. at t = 4.0

prohojiy [21]

X =(3.00x4.00 x3-1.00t x 2.00) x m

x= (12.00x3- 1.00 x2.00) x m
x= 36.00 -1.00 x 2.00) x m
x = (36.00 -2.00) x m
x =( 34.00) x m
x =34.00 times m

7 0

3 years ago