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1 year ago

Picking up sound waves is the stage of listening that most people refer to as what?

Physics

1 answer:

kirill [66]1 year ago

5 0

A. Picking up sound waves is the stage of listening that most people refer to as hearing.

<h3>What is hearing?</h3>

Hearing, or auditory perception, is the ability to perceive sounds through an organ, such as an ear, by detecting vibrations as periodic changes in the pressure of a surrounding medium.

During learning process, when someone picks up the sound waves directed to him/her, it is known as the act of hearing,

Thus, picking up sound waves is the stage of listening that most people refer to as hearing.

Learn more about hearing here: brainly.com/question/14185060

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A drop of oil of volume 10m it spread out on water to make a circular firm of radius 10m calculate the tickness of the firm

Effectus [21]

Answer:

h = 3.1 cm

Explanation:

Given that,

The volume of a oil drop, V = 10 m

Radius, r = 10 m

We need to find the thickness of the film. The film is in the form of a cylinder whose volume is as follows :

$V=\pi r^2 h\\\\h=\dfrac{V}{\pi r^2}\\\\h=\dfrac{10}{\pi \times 10^2}\\\\h=0.031\ m\\\\h=3.1\ cm$

So, the thickness of the film is equal to 3.1 cm.

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2 years ago

A ball is thrown toward a cliff of height h with a speed of 33 m/s and an angle of 60∘ above horizontal. It lands on the edge of

Veseljchak [2.6K]

Answer:

(a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

Explanation:

Given that,

Speed = 33 m/s

Angle = 60°

Time = 3.0 sec

(a). We need to calculate the height of the cliff

Using equation of motion

$h=ut-\dfrac{1}{2}gt^2$

$h=(u\sin60)\times t-\dfrac{1}{2}gt^2$

Put the value into the formula

$h=33\times\sin60\times3.0-\dfrac{1}{2}\times9.8\times(3.0)^2$

$h=41.6\ m$

(b). We need to calculate the maximum height of the ball

Using formula of height

$h_{max}=\dfrac{(u\sin\theta)^2}{2g}$

Put the value into the formula

$h=\dfrac{(33\sin60)^2}{2\times 9.8}$

$h=41.67\ m$

(c). We need to calculate the vertical component of velocity of ball

Using equation of motion

$v=u-gt$

$v=u\sin\theta-gt$

Put the value into the formula

$v_{y}=33\times\sin 60-9.8\times3.0$

$v_{y}=-0.82\ m/s$

We need to calculate the horizontal component of velocity of ball

Using formula of velocity

$v_{x}=u\cos\theta$

Put the value into the formula

$v_{x}=33\times\cos60$

$v_{x}=16.5\ m/s$

We need to calculate the ball's impact speed

Using formula of velocity

$v=\sqrt{v_{x}^2+v_{y}^2}$

Put the value into the formula

$v=\sqrt{(16.5)^2+(-0.82)^2}$

$v=16.52\ m/s$

Hence, (a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

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torisob [31]

Answer:

yes

Explanation:

Let's solve your equation step-by-step.

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Step 1: Subtract 3 from both sides.

4x+3−3=−5−3

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Step 2: Divide both sides by 4.

4x / 4 = −8 / 4

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Answer: the answer is a

Explanation:

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