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3 years ago

5

Having established that a sound wave corresponds to pressure fluctuations in the medium, what can you conclude about the directi

on in which such pressure fluctuations travel

1 answer:

DaniilM [7]3 years ago

4 0

Answer:

the pressure fluctuation is LONGITUDINAL

Explanation:

Sound waves are an oscillating movement of air particles, this can be analyzed in two different, but equivalent ways, as an air oscillation and with a pressure wave due to these oscillations.

The expression for the wave is

ΔP = Δo sin (kx - wt)

Therefore, the pressure variation is in the same direction as the displacement variation, consequently the pressure fluctuation is LONGITUDINAL

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If something is being carried by westerlies, what direction is it moving??

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The train slows down at the railroad crossing. Is that acceleration, velocity, speed, of none of the above?

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3 years ago

A cart is driven by a large propeller or fan, which can accelerate or decelerate the cart. The cart starts out at the position x

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Answer:

The acceleration of the cart is 1.0 m\s^2 in the negative direction.

Explanation:

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5 0

3 years ago

The basic barometer can be used to measure the height of a building. If the barometric readings at the top and at the bottom of

Elan Coil [88]

Answer: 230.50 m

Explanation:

We have the following information:

$h_{Hg-TOP}=675mmHg=0.675m$ the barometric reading at the top of the building

$h_{Hg-BOT}=695mmHg=0.695m$ the barometric reading at the bottom of the building

$\rho _{air}=1.18 kg/m^{3}$ density of air

$\rho _{Hg}=13600 kg/m^{3}$ density of mercury

$g=9.8/m^{2}$ gravity

And we need to find the height of the building.

In order to approach this problem, we will firstly use the following equations to find the pressure at the top of the building $P_{TOP}$ and the perssure at the bottom $P_{BOT}$ :

$P_{TOP}=\rho _{Hg} g h_{Hg-TOP}$ (1)

$P_{BOT}=\rho _{Hg} g h_{Hg-BOT}$ (2)

From (1): $P_{TOP}=(13600 kg/m^{3})(9.8/m^{2})(0.675m)=89964 Pa$ (3)

From (2): $P_{BOT}=(13600 kg/m^{3})(9.8/m^{2})(0.695m)=92629.6 Pa$ (4)

Having the pressures at the top and the bottom of the building, we can calculate the variation in pressure $\Delta P$ :

$\Delta P=P_{BOT} - P_{TOP}$ (5)

$\Delta P=92629.6 Pa - 89964 Pa=2665.6 Pa$ (6)

On the other hand, we have a column of air with a cross-section area $A$ and the same height of the building, lets name it $h_{air}$ .

As pressure is defined as the force $F$ exerted on a specific area $A$ , we can write:

$\Delta P=\frac{F}{A}$ (7)

If we isolate $F$ we have:

$F= A \Delta P$ (8)

Also, the force gravity exerts on this column of air (its weight) is:

$F=m_{air} g$ (9)

Knowing the density of air is: $\rho_{air}=\frac{m_{air}}{V_{air}}$ (10)

where the volume of air can be written as: $V_{air}=(A)(h_{air})$ (11)

Substituting (1) in (10):

$\rho_{air}=\frac{m_{air}}{(A)(h_{air}}$ (12)

Isolating $m_{air}$ :

$m_{air}=(\rho_{air}) (A) (h_{air})$ (13)

Substituting (13) in (9):

$F=(\rho_{air}) (A) (h_{air}) (g)$ (14)

Matching (8) and (14)

$A \Delta P=(\rho_{air}) (A) (h_{air}) (g)$ (15)

Isolating $h_{air}$ :

$h_{air}=\frac{\Delta P}{g \rho_{air}}$ (16)

Substituting the known and calculated values:

$h_{air}=\frac{2665.6 Pa}{(9.8m/s^{2}) (1.18 kg/m^{3})}$ (17)

Finally:

$h_{air}=230.50 m$ This is the height of the building

8 0

3 years ago