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tiny-mole [99]
3 years ago
15

When responding to sound, the human eardrum vibrates about its equilibrium position. Suppose an eardrum is vibrating with an amp

litude of 7.2 x10-7 m and a maximum speed of 3.6 x10-3 m/s. (a) What is the frequency (in Hz) of the eardrum's vibrations
Physics
1 answer:
Paraphin [41]3 years ago
6 0

Answer:

796.18 Hz

Explanation:

Applying,

Maximum velocity = Amplitude×Angular velocity

Therefore,

V' = A(2πf)............... Equation 1

Where V' = maximum velocity of the eardrum, A = Amplitude of vibration of the eardrum, f = frequency of the eardrum vibration, π = pie

make f the subject of the equation

f = V'/2πA................ Equation 2

From the question,

Given: V' = 3.6×10⁻³ m/s, A' = 7.2×10⁻⁷ m,

Constant: 3.14.

Substitute these values into equation 2

f = 3.6×10⁻³/( 7.2×10⁻⁷×2×3.14)

f = 796.18 Hz

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The era of planet formation ended when the remaining hydrogen and helium gas of the solar nebula was swept into interstellar spa
artcher [175]

Answer:

The era of planet formation ended when the remaining hydrogen and helium gas of the solar nebula was swept into interstellar space by the solar winds.

Explanation:          

The Solar System is formed from a molecular cloud (compound by gas and dust). If there is a near perturbation to the cloud, maybe due to a supernova explosion, the molecular cloud will collapse under its own gravity. Then, in some point it starts to rotate and will accrete all the material in a disk around the protostar¹.

Inside the disk, dust particles start to collide and accrete until they form planetesimals². As a consequence of the gravitational force of the star, rocky and metallic particles will be more attracted to the inner part of the Solar System (close to the Sun) since they have more mass than gas.

Then, when the star has the necessary pressure and temperature to initiate nuclear reactions in its core, it will be able to emit huge amounts of energy, better known as solar winds. These winds will expel gas (hydrogen and helium) from the Solar System more easily than the rocky and metallic  particles.

Notice that when such event occurs, rocky and gaseous planets were already formed.

       

Key terms:

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5 0
3 years ago
How much time does it take for an eagle flying at a speed of 50 kilometers per hour to travel a distance of 2000 kilometers?
gtnhenbr [62]

Answer:40 hour

Explanation:

8 0
3 years ago
Suppose a baseball pitcher throws the ball to his catcher.
amm1812

a) Same

b) Same

c) Same

d) Throw the ball takes longer

e) F is larger when the ball is catched

Explanation:

a)

The change in speed of an object is given by:

\Delta v = |v-u|

where

u is the initial velocity of the object

v is the final velocity of the object

The change in speed is basically the magnitude of the change in velocity (because velocity is a vector, while speed is a scalar, so it has no direction).

In this problem:

- In situation 1 (pitcher throwing the ball), the initial velocity is

u = 0 (because the ball starts from rest)

while the final velocity is v, so the change in speed is

\Delta v=|v-0|=|v|

- In situation 2 (catcher receiving the ball), the initial velocity is now

u = v

while the final velocity is now zero (ball coming to rest), so the change in speed is

\Delta v =|0-v|=|-v|

Which means that the two situations have same change in speed.

b)

The change in momentum of an object is given by

\Delta p = m \Delta v

where

m is the mass of the object

\Delta v is the change in velocity

If we want to compare only the magnitude of the change in momentum of the object, then it is given by

|\Delta p|=m|\Delta v|

- In situation 1 (pitcher throwing the ball), the change in momentum is

\Delta p = m|\Delta v|=m|v|=mv

- In situation 2 (catcher receiving the ball), the change in momentum is

\Delta p = m\Delta v = m|-v|=mv

So, the magnitude of the change in momentum is the same (but the direction is opposite)

c)

The impulse exerted on an object is equal to the change in momentum of the object:

I=\Delta p

where

I is the impulse

\Delta p is the change in momentum

As we saw in part b), the change in momentum of the ball in the two situations is the same, therefore the impulse exerted on the ball will also be the same, in magnitude.

However, the direction will be opposite, as the change in momentum has opposite direction in the two situations.

d)

To compare the time of impact in the two situations, we have to look closer into them.

- When the ball is thrown, the hand "moves together" with the ball, from back to ahead in order to give it the necessary push. We can verify therefore that the time is longer in this case.

- When the ball is cacthed, the hand remains more or less "at rest", it  doesn't move much, so the collision lasts much less than the previous situation.

Therefore, we can say that the time of impact is longer when the ball is thrown, compared to when it is catched.

e)

The impulse exerted on an object can also be rewritten as the product between the force applied on the object and the time of impact:

I=F\Delta t

where

I is the impulse

F is the force applied

\Delta t is the time of impact

This can be rewritten as

F=\frac{I}{\Delta t}

In this problem, in the two situations,

- I (the impulse) is the same in both situations

- \Delta t when the ball is thrown is larger than when it is catched

Therefore, since F is inversely proportional to \Delta t, this means that the force is larger when the ball is catched.

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It takes the shape of the cup and it can be sucked through a straw 
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