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stellarik [79]
3 years ago
14

If a fast-moving car making a loud noise approaches and moves past the person, what will happen as the distance between the two

increases
which one is it?
A. The pitch of the sound being heard by the person will appear to be higher than the pitch of the source.
B. The frequency of the sound waves reaching the person's ear will be greater than the frequency of the waves leaving the car.
 C. The pitch and frequency of the sound waves reaching the person's ear will remain unchanged.
D. The pitch of the sound being heard by the person will appear to be lower than the pitch of the source.  
Chemistry
2 answers:
Pani-rosa [81]3 years ago
5 0

Correct answer choice is :


D) The pitch of the sound being heard by the person will appear to be lower than the pitch of the source.


Explanation:


The sensation of a frequency is usually related to as the pitch of a sound. A high pitch sound relates to a high-frequency sound wave and a low pitch sound relates to a low-frequency sound wave. Whereas loudness depends on the power of the wave. In overall, the pitch is the cause behind the variation in voice quality for several people. The pitch of a sound depends on the frequency while loudness of a sound depends on the amplitude of sound waves.

wlad13 [49]3 years ago
4 0
If a fast-moving car making a loud noise approaches and moves past the person, according to the Doppler effect, what will happen as the distance between the two increases is that D) the pitch of the sound being heard by the person will appear to be lower than the pitch of the source. 
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natka813 [3]

Answer:

The light is most intense where it strikes Earth _ directly_______ to its surface.

Explanation:

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3 0
3 years ago
Read 2 more answers
20cm of 0.09M solution of H2SO4. requires 30cm of NaOH for complete neutralization. Calculate the
kirill115 [55]

Answer:

Choice A: approximately 0.12\; \rm M.

Explanation:

Note that the unit of concentration, \rm M, typically refers to moles per liter (that is: 1\; \rm M = 1\; \rm mol\cdot L^{-1}.)

On the other hand, the volume of the two solutions in this question are apparently given in \rm cm^3, which is the same as \rm mL (that is: 1\; \rm cm^{3} = 1\; \rm mL.) Convert the unit of volume to liters:

  • V(\mathrm{H_2SO_4}) = 20\; \rm cm^{3} = 20 \times 10^{-3}\; \rm L = 0.02\; \rm L.
  • V(\mathrm{NaOH}) = 30\; \rm cm^{3} = 30 \times 10^{-3}\; \rm L = 0.03\; \rm L.

Calculate the number of moles of \rm H_2SO_4 formula units in that 0.02\; \rm L of the 0.09\; \rm M solution:

\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V(\mathrm{H_2SO_4})\\ &= 0.02 \; \rm L \times 0.09 \; \rm mol\cdot L^{-1} = 0.0018\; \rm mol \end{aligned}.

Note that \rm H_2SO_4 (sulfuric acid) is a diprotic acid. When one mole of \rm H_2SO_4 completely dissolves in water, two moles of \rm H^{+} ions will be released.

On the other hand, \rm NaOH (sodium hydroxide) is a monoprotic base. When one mole of \rm NaOH formula units completely dissolve in water, only one mole of \rm OH^{-} ions will be released.

\rm H^{+} ions and \rm OH^{-} ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid \rm H_2SO_4 dissolves in water completely, it will take two moles of \rm OH^{-} to neutralize that two moles of \rm H^{+} produced. On the other hand, two moles formula units of the monoprotic base \rm NaOH will be required to produce that two moles of \rm OH^{-}. Therefore, \rm NaOH and \rm H_2SO_4 formula units would neutralize each other at a two-to-one ratio.

\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\; H_2O.

\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2.

Previous calculations show that 0.0018\; \rm mol of \rm H_2SO_4 was produced. Calculate the number of moles of \rm NaOH formula units required to neutralize that

\begin{aligned}n(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})}\cdot n(\mathrm{H_2SO_4}) \\&= 2 \times 0.0018\; \rm mol = 0.0036\; \rm mol\end{aligned}.

Calculate the concentration of a 0.03\; \rm L solution that contains exactly 0.0036\; \rm mol of \rm NaOH formula units:

\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} = \frac{0.0036\; \rm mol}{0.03\; \rm L} = 0.12\; \rm mol \cdot L^{-1}\end{aligned}.

3 0
3 years ago
acetaminophen is a stronger acid than methanol, and sodium methoxide is a stronger base than. write an acid base equilibrium rea
torisob [31]

Answer:

The equilibrium shifts to the right that is the forward reaction.

Explanation:

The chemical compound known as "Acetaminophen" is a chemical compound that is generally known to a layman as Paracetamol and it belongs to the drug class known as anagelsics which helps in the treatment of pain or say in the reduction of pain. Acetaminophen has the chemical Formula to be C8H9NO2, with the Molar mass of 151.163 g/mol and Boiling point of 420 °C.

The reaction between Acetaminophen and sodium methoxide gives methanol and acetaminophen sodium salt. Therefore, the acid base equilibrium reaction of these species is given as;

C8H9NO2 + CH3ONa <========> CH3OH + acetaminophen sodium salt.

The equilibrium shifts to the right that is the forward reaction.

8 0
3 years ago
The activation energy of an uncatalyzed reaction is 95kJ/mol. The addition of a catalyst lowers the activation energy to 55kJ/mo
notka56 [123]

Answer:

a) at 25°C the rate of reaction increases by a factor of 1,027*10^7

b) at 25°C the rate of reaction increases by a factor of 1,777*10^5

Explanation:

using the Arrhenius equation

k= ko*e^(-Ea/RT)

where

k= reaction rate

ko= collision factor

Ea= activation energy

R= ideal gas constant= 8.314 J/mol*K

T= absolute temperature

for the uncatalysed reaction

k1= ko*e^(-Ea1/RT)

for the catalysed reaction

k2= ko*e^(-Ea2/RT)

dividing both equations

k2/k1= e^(-(Ea2-Ea1)/RT)

a) at 25°C

k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,027*10^7

therefore at 25°C , k2/k1 = 1,027*10^6

b) at 125°C

k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,777*10^5

therefore at 125°C , k2/k1 = 1,777*10^5

Note:

when the catalysts is incorporated, the catalysed reaction and the uncatalysed one run in parallel and therefore the real reaction rate is

k real = k1 + k2 = k2 (1+k1/k2)

since k2>>k1 → 1+k1/k2 ≈ 1 and thus k real ≈ k2

6 0
3 years ago
how much current would be measured in a circuit if the light bulb has a resistance of 6 ohms and a voltage of 36 volts
Stels [109]

Answer:

The right response is "6 A". A further explanation is given below.

Explanation:

The given values are:

Resistance,

R = 6 ohms

Voltage,

V = 36 volts

As we know,

⇒  V=IR

then,

⇒  I=\frac{V}{R}

On substituting the values, we get

⇒     =\frac{36}{6}

⇒     =6 \ A

8 0
3 years ago
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