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3 years ago

Does sound travels through your skull to your ear when you speak ? If yes then how ?

1 answer:

5 0

Sound travels through waves, more specifically, through vibrations. They do not go from skull to ear, but they can go from ear to brain, or skull to brain. Ear to brain is simply vibrations traveling from outer ear, to inner ear, to the brain. Skull to brain, otherwise known as "bone conduction", has the vibrations hitting the skull, then to the temproal bone, then to the inner ear where the brain picks it up.

You might be interested in

The skin contains touch receptors to respond to stimuli.

AURORKA [14]

Answer:

true

Explanation:

receptors that are left in the body sense touch ar located ub tge two layer of the skin the demis ane epidermis.

5 0

3 years ago

A planet is discovered orbiting around a star in the galaxy Andromeda at the same distance from the star as Earth is from the Su

aniked [119]

Answer:

The planet´s orbital period will be one-half Earth´s orbital period.

Explanation:

The planet in orbit, is subject to the attractive force from the sun, which is given by the Newton´s Universal Law of Gravitation.

At the same time, this force, is the same centripetal force, that keeps the planet in orbit (assuming to be circular), so we can put the following equation:

Fg = Fc ⇒ G*mp*ms / r² = mp*ω²*r

As we know to find out the orbital period, as it is the time needed to give a complete revolution around the sun, we can say this:

ω = 2*π / T (rad/sec), so replacing this in the expression above, we get:

Fg = Fc ⇒ G*mp*ms / r² = mp*(2*π/T)²*r

Solving for T²:

T² = (2*π)²*r³ / G*ms (1)

For the planet orbiting the sun in Andromeda, we have:

Ta² = (2*π)*r³ / G*4*ms (2)

As the radius of the orbit (distance to the sun) is the same for both planets, we can simplify it in the expression, so, if we divide both sides in (1) and (2), simplifying common terms, we finally get:

(Te / Ta)² = 4 ⇒ Te / Ta = 2 ⇒ Ta = Te/2

So, The planet's orbital period will be one-half Earth's orbital period.

7 0

3 years ago

At which point should a researcher inform the respondents about the use of deception in a psychological study ??

lbvjy [14]

D during the course of the study

8 0

3 years ago

An air bubble at the bottom of a lake 36.0 m deep has a volume of 1.22 cm^3. If the temperature at the bottom is 5.9°C and at th

AlexFokin [52]

Answer:

volume of the bubble just before it reaches the surface is 5.71 cm³

Explanation:

given data

depth h = 36 m

volume v2 = 1.22 cm³ = 1.22 × $10^{-6}$ m³

temperature bottom t2 = 5.9°C = 278.9 K

temperature top t1 = 16.0°C = 289 K

to find out

what is the volume of the bubble just before it reaches the surface

solution

we know at top atmospheric pressure is about P1 = $10^{5}$ Pa

so pressure at bottom P2 = pressure at top + ρ×g×h

here ρ is density and h is height and g is 9.8 m/s²

pressure at bottom P2 = $10^{5}$ + 1000 × 9.8 ×36

pressure at bottom P2 =4.52 × $10^{5}$ Pa

so from gas law

$\frac{P1*V1}{t1} = \frac{P2*V2}{t2}$

here p is pressure and v is volume and t is temperature

so put here value and find v1

$\frac{10^{5}*V1}{289} = \frac{4.52*10^{5}*1.22}{278.9}$

V1 = 5.71 cm³

volume of the bubble just before it reaches the surface is 5.71 cm³

6 0

3 years ago

How much work is done by an applied force to lift a 45 newton block 6.0 meters at a constant speed ?

AleksandrR [38]

Answer:

270Joues

Explanation:

Step one:

given data

Force F= 45N

distance moved d= 6m

Required

The work done in moving the block 6m

Step two:

We know that the expression for the work done is

WD= force* distance

WD= 45*6

WD=270Joues

7 0

3 years ago