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3 years ago

10

You are standing on a street corner and you hear an ambulance siren. The pitch of the siren seems to be getting higher. What do

you expect will happen next?
A. The pitch of the siren will change unpredictably as it approaches you.
B. The siren will get louder and higher as the ambulance moves towards you.
C. The siren will eventually fade as the ambulance moves away from you.
D. The pitch of the siren will continue to get higher, but you will not see the ambulance because it is moving away.

2 answers:

Ira Lisetskai [31]3 years ago

8 0

The answer is B

I know this because if the siren seems to be getting higher that means that the Ambulance is coming towards you. Which is was answer b is saying.

Hope this helps please make me brainliest.

Gre4nikov [31]3 years ago

4 0

B.The siren will get louder and higher as the ambulance moves towards you

Hope this helped !

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vladimir2022 [97]

Answer:

S= 1.40x10⁻⁵mol/L

Explanation:

The Henry's Law is given by the next expression:

$S = k_{H} \cdot p$ (1)

<em>where S: is the solubility or concentration of Ar in water, $k_{H}$ : is Henry's law constant and p: is the pressure of the Ar </em>

<u>Since the argon is 0.93%, we need to multiply the equation (1) by this percent:</u>

$S = 1.5 \cdot 10^{-3} \frac{mol}{L\cdot atm} \cdot 1.0atm \cdot \frac{0.93}{100} = 1.40 \cdot 10^{-5} \frac{mol}{L}$

Therefore, the argon solubility in water is 1.40x10⁻⁵mol/L.

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Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.

DiKsa [7]

Answer:0.061

Explanation:

Given

$T_C=300 k$

Temperature of soup $T_H=340 K$

heat capacity of soup $c_v=33 J/K$

Here Temperature of soup is constantly decreasing

suppose T is the temperature of soup at any instant

efficiency is given by

$\eta =\frac{dW}{Q}=1-\frac{T_C}{T}$

$dW=Q(1-\frac{T_C}{T})$

$dW=c_v(1-\frac{T_C}{T})dT$

integrating From $T_H$ to $T_C$

$\int dW=\int_{T_C}^{T_H}c_v(1-\frac{T_C}{T})dT$

$W=\int_{T_C}^{T_H}33\cdot (1-\frac{300}{T})dT$

$W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}$

$W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]$

Now heat lost by soup is given by

$Q=c_v(T_C-T_H)$

Fraction of the total heat that is lost by the soup can be turned is given by

$=\frac{W}{Q}$

$=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}$

$=\frac{T_C-T_H-T_C\ln (\frac{T_C}{T_H})}{T_C-T_H}$

$=\frac{300-340-300\ln (\frac{300}{340})}{300-340}$

$=\frac{-40+37.548}{-40}$

$=0.061$

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sweet [91]

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27,000cm^3 * 1g/cm^3 = 27,000 grams = 27 kg = Mass of water displaced.

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3 years ago