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2 years ago

15

boat travels through a river at a speed of 25 m/s, passing through schools and opening the soundtrack. One student measured freq

uency 610 Hz. Turns out, four seconds later, the student again measured frequency 587 Hz if the river was a straight line 20 m wide and the speed of sound is 340 m/s. find school distance from coast?

2 answers:

anyanavicka [17]2 years ago

4 0

Answer:

256782

Explanation:302-320+3223

True [87]2 years ago

4 0

302-320+3223 maybbbb

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The operating coefficient or performance coefficient of a heat pump is the ratio between the heating or cooling provided and the electricity consumed. The higher coefficients are equivalent to lower operating costs. The coefficient can be greater than 1, because it is a percentage of the output: losses, other than the thermal efficiency ratio: input energy. Mathematically can be written as,

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3 years ago

an airplane traveling 245 m/s east expericences turbulence, so the pilot slows down to 230 m/s. it takes the pilot 7 seconds to

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3 years ago

The position of the front bumper of a test car under microprocessor control is given by x(t) = 2.31 m + (4.90 m/s2)t2 - (0.100 m

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The equation of the car is given by the equation,

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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