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3 years ago

How does being underwater change the ability to localize sound? why is this so?

1 answer:

3 0

Being underwater REDUCES the ability to localize sound. This is because, the resonance frequency of the external ear is reduced when the outer ear canal is fill with water. The presence of the water - air interface when one is under water and the reduction in the impedance matching ability of the middle ear also contribute to the reduction in one's ability to localize sound underwater.

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Two objects, with different sizes, masses, and temperatures, are placed in thermal contact. in which direction does the energy t

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<span>(c) energy travels from the object at higher temperature
to the object at lower temperature.

Size and mass have no effect.</span>

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3 years ago

What is a substance that when mixed with water turns blue litmus paper red

alekssr [168]

When blue litmus paper is dipped in acid the paper turns red.

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3 years ago

PICTURE ABOVE !

Rom4ik [11]

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she should ve the one handing out the cookies each day, to make sure each child gets only one cookie a day.

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3 years ago

6. A resident throws a bag full of trash over the balcony of their third-floor apartment. What will happen to the acceleration o

trasher [3.6K]

The trash bag will gain gain speed in till hitting the ground and passably exploding on impact of the ground depending on the strength of the trash bag it may rip during the fall.

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6 0

3 years ago

Water drips from the nozzle of a shower onto the floor 193 cm below. The drops fall at regular (equal) intervals of time, the fi

Law Incorporation [45]

Answer: 108.81 cm and 48.66 cm

Explanation:

In this, we have to make sure to keep in mind the Gravity effects on the drops. The drops will accelerate when they fall making them travel faster. This means, the velocity is not constant.

What is know:

Height (h) = 193 cm

Gravity (g) = 981 $cm/s^{2}$

Initial Velocity = 0

First, we can know how long it take to the drop to travel to the floor. It can be done with the following equation:

$x = V_{0} t + \frac{1}{2} at^{2}$ (1)

Where:

x is the distance which is 193cm

Vo is the Initial Velocity which is zero

t is the time the time it takes the drop to travel from the shower to the floor

a is the aceleration, which in this case is the gravity.

With the Initial Velocity equals zero the equations simply:

$193 cm = \frac{1}{2}gt^{2}$

To search for the time:

$t =\sqrt{\frac{2*193cm}{981cm/s^{2} } }$

t = 0.627 s

This is the time it takes a drop to fall to the floor, with this time and knowing other 3 drops have driped from the shower by this time. We can calculate how much time it takes the shower to drip each drop.

Time for Drip = t/4

Time for Drip = 0.156

This time is the difference between each drop, using the same equation we can calculate where was each drop, because now it is know how much time had each drop after being drip from the shower.

Our first is already on the floor (193 cm) with 0.627 s, The second drops have been falling for (0.627s - 0.156) 0.471 s and our third drop for (0.627s - 0.156 - 0.156) 0.315 s

We can use (1) to know how far have each drop traveled on these times. We know the Initials Velocity are 0, know we need ot know the distances.

For the second drop:

$x = \frac{1}{2} (981cm/s^{2})(0.471s)^{2}$

$x =108.81 cm$

For the third drop:

$x = \frac{1}{2} (981cm/s^{2})(0.315s)^{2}$

$x = 48.66 cm$

8 0

2 years ago