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2 years ago

How does the ringing sound of a telephone travels from the phone to your ear

Physics

2 answers:

lilavasa [31]2 years ago

8 0

Answer:

Sound produced by a vibrating object reaches our ear through sound waves which propagate through the medium as a series of compression and rarefaction.

Auditory receptors receptors receive vibrations from telephone and they are passed to ear drum which then passes it auditory ossicles which then passes it to fluid of cochlea which then basilar membrane and then finally it reaches to hair cells. Then the sound is heard

AysviL [449]2 years ago

5 0

Sound waves travel through the air to your eardrum

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What is the height of the Golden Bridge?

Mademuasel [1]

Answer:

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Explanation:

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2 years ago

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NemiM [27]

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Explanation:

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2 years ago

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2 years ago

There is a limit to how long your neck can be. If your neck were too long, no blood would reach your brain! What is the maximum

spayn [35]

Answer:

The maximum height a person's brain could be above his heart is: 1.28 meter.

Explanation:

We need to know what is the normal blood pressure ours hearts so there is a rate: 120/80 (mmHg) and the average will be: 100 (mmHg) and using the Pascal law that relate pressure, density, gravity and height like: $P_{2} = pgh_{1} - pgh_{2} + P_{1}$ , where P is pressure, p is density, g is the gravity acceleration and h is the height. Now we can find the height and delta of pressure will be: P2-P1 = 100 (mmHg), knowing that 1(mmHg) is equal to 133 Pa, we can do the convertion to 13332.2 (Pa), now because the units of Pascal are Newton/(meter^2). Then we solve the formula to get the height: $\frac{P2-P1}{pg} =h$ so we get: $\frac{13332.2}{(1060*9,81)}=Height=1.28(meters).$

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3 years ago

A kangaroo can jump over an object 2.46 m high. (a) Calculate its vertical speed (in m/s) when it leaves the ground.

Elenna [48]

Answer:

a) 6.95 m/s

b) 1.42 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 2.46-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 2.46}\\\Rightarrow u=6.95\ m/s$

a) The vertical speed when it leaves the ground. is 6.95 m/s

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-6.95}{-9.81}\\\Rightarrow t=0.71\ s$

Time taken to reach the maximum height is 0.71 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2.46=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.46\times 2}{9.81}}\\\Rightarrow t=0.71\ s$

Time taken to reach the ground from the maximum height is 0.71 seconds

b) Time it stayed in the air is 0.71+0.71 = 1.42 seconds

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3 years ago