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3 years ago

Witch part of the ear is responsible for sending singnals about sound to the brain

Physics

1 answer:

drek231 [11]3 years ago

7 0

Answer:

Cochlea

Explanation:

Cochlea. The $cochlea$ changes the sound vibrations from the middle ear into nerve signals. The nerve signals the travel to the brain via the $cochlea \ nerve$ .

The cochlea nerve is also called the auditory nerve.

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Two circus performers rehearse a trick in which a ball and a dart collide. Horatio stands on a platform 9.8 m above the ground a

xenn [34]

Answer:

Time = 0.55 s

Height = 8.3 m

Explanation:

The ball is dropped and therefore has an initial velocity of 0. Its acceleration, g, is directed downward in the same direction as its displacement, $h_b$ .

The dart is thrown up in which case acceleration, g, acts downward in an opposite direction to its displacement, $h_d$ . Both collide after travelling for a time period, t. Let the height of the dart from the ground at collision be $h_d$ and the distance travelled by the ball measured from the top be $h_b$ .

It follows that $h_d+h_b=9.8$ .

Applying the equation of motion to each body (h = v_0t + 0.5at^2),

Ball:

$h_b=0\times t + 0.5\times 9.8t^2$ (since $v_{b0} =0$ .)

$h_b=4.9t^2$

Dart:

$h_d=17.8\times t - 0.5\times9.8t^2$ (the acceleration is opposite to the displacement, hence the negative sign)

$h_d=17.8\times t - 4.9t^2$

But

$h_b+h_d =9.8$

$17.8\times t - 4.9t^2+4.9t^2 =9.8$

$17.8\times t = 9.8$

$t = 0.55$

The height of the collision is the height of the dart above the ground, $h_d$ .

$h_d=17.8\times t - 4.9t^2$

$h_d=17.8\times 0.55 - 4.9\times(0.55)^2$

$h_d=9.79 - 1.48225$

$h_d=8.3$

8 0

4 years ago

WILL GIVE BRAINLIEST!!!!!!

DENIUS [597]

Answer: option d: The nucleus of Atom Q is more stable than the nucleus of Atom P.

Explanation:

Atom P is radioactive and disintegrates, it emits beta particles (high speed electrons or positrons) because it is not stable. On disintegration, it forms a stable Atom Q which is non-radioactive and thus it does not disintegrates further.

Thus, the correct option is only d. The nucleus of Atom Q is more stable than the nucleus of Atom P.

8 0

3 years ago

What temperature increase is necessary to increase the power radiated from an object by a factor of 8?

Aleks04 [339]

To solve this problem it is necessary to apply the concepts related to the Power defined from the Stefan-Boltzmann equations.

The power can be determined as:

$P = \sigma T^4$

Making the relationship for two states we have to

$\frac{P_1}{P_2} = \frac{T_1^4}{T_2^4}$

Since the final power is 8 times the initial power then

$P_2 = 8P_1$

Substituting,

$\frac{1}{8} = \frac{T_1^4}{T_2^4}$

$T_2 = T_1 8*(\frac{1}{4})$

$T_2 = 1,68T_1$

The temperature increase would then be subject to

$\Delta T = T_2-T_1$

$\Delta T = 1.68T_1 -T_1$

$\Delta T = 0.68T_1$

The correct option is D, about 68%

7 0

3 years ago

If a 1.0-kg-mass block is on the left cap, how much total mass must be placed on the right cap so that the caps equilibrate at e

Dennis_Churaev [7]

<span>The total mass that should be placed in the right cap so that the caps equilibrate at equal height is also 1 kg. if equilibrium should be maintained the force in each side should cancel out, so to balance a 1kg mass, a 1 kg mass should also be place on the opposite direction</span>

3 0

3 years ago

it takes a hose 4 minutes to fill a rectangular aquarium 9 inches long, 13 inces wide, and 14 inces tall. how long will it take

Studentka2010 [4]

Answer: 43.58 min

Explanation:

Knowing the volume of a rectangular object is length x width x height, we have two volumes:

$V_{1}=(9 in)(13 in)(14 in)=1638 in^{3}$

$V_{2}=(21 in)(25 in)(34 in)=17850 in^{3}$

And we know it takes a time of 4 minutes to fill $V_{1}$ .

If we want to know how long will it take the same hose to fill another tank with volume $V_{2}$ , we can use the <u>Rule of three</u>, which is a mathematical rule to find out an amount that is with another quantity given in the same relation as other two also known:

$1638in^{3}$ ---- $4min$

$17850in^{3}$ ---- $?$

$?=\frac{(17850 in^{3})(4min)}{1638 in^{3}}$

Finally:

$?=43.58 min$

4 0

4 years ago