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GrogVix [38]
3 years ago
10

What is the frequency of a wave traveling at 5,000 m/s with a wavelength of .002 m/cycle?

Chemistry
1 answer:
Korvikt [17]3 years ago
8 0

Answer:

2.5\cdot 10^6 Hz

Explanation:

The relationship between the speed of a wave, its frequency and its wavelength is given by the wave equation:

v=f\lambda

where

v is the wave speed

f is the frequency of the wave, which is the number of cycles completed by the wave in one oscillation

\lambda is the wavelength, which is the distance between two consecutive crests of the wave

For the wave in this problem we have:

v = 5000 m/s is the speed

\lambda=0.002 m is the wavelength of the wave

Solving for f, we find the frequency:

f=\frac{v}{\lambda}=\frac{5000}{0.002}=2.5\cdot 10^6 Hz

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Answer:

0.39 mol

Explanation:

Considering the ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

At same volume, for two situations, the above equation can be written as:-

\frac {{n_1}\times {T_1}}{P_1}=\frac {{n_2}\times {T_2}}{P_2}

Given ,  

n₁ = 1.50 mol

n₂ = ?

P₁ = 3.75 atm

P₂ = 0.998 atm

T₁ = 21.7  ºC

T₂ = 28.1 ºC

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (21.7 + 273.15) K = 294.85 K  

T₂ = (28.1 + 273.15) K = 301.25 K  

Using above equation as:

\frac{{n_1}\times {T_1}}{P_1}=\frac{{n_2}\times {T_2}}{P_2}

\frac{{1.50\ mol}\times {294.85\ K }}{3.75\ atm}=\frac{{n_2}\times {301.25\ K  }}{0.998\ atm}

n_2=\frac{{1.50}\times {294.85}\times 0.998}{3.75\times 301.25}\ mol

Solving for n₂ , we get:

n₂ = 0.39 mol

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Why is the Hubble telescope located in space? *
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Explanation:

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3 years ago
You decide to clean the bathroom. You notice that the shower is covered in a strange green slime . you try to get rid of this sl
Free_Kalibri [48]
  1. The independent variable (IV) is the lemon juice mixture
  2. The dependent variable (DV) is the appearance of the green slime on the shower
  3. The control variable (CV) are time taken to spray, the amount of spray
  4. The experimental group (EG) is the side of the shower sprayed with lemon juice mixture
  5. The control group (CG) is the side of the shower sprayed with water.

INDEPENDENT VARIABLE

  • Independent variable is the variable of an experiment that is changed by the experimenter in order to bring about a change. It is the variable being tested in the experiment. In this case, the IV is the lemon juice mixture tested on the green slime on the shower.

DEPENDENT VARIABLE:

  • Dependent variable is the variable that is observed or measured in an experiment. It is also called responding variable. The DV in this case is the appearance of the green slime on the shower.

CONTROL VARIABLE:

  • Control variable is the variable that is kept constant throughout the experiment for all groups. The CV is the same for all the groups and they include: time taken to spray, the same amount of spray

CONTROL GROUP

  • Control group is the group that does not receive the independent variable or test in an experiment. In this case, the CG is the side of the shower sprayed with water.

EXPERIMENTAL GROUP:

  • Experimental group is the group of ab experiment that receives the experimental treatment or independent variable. In this case, the EG is the side of the shower sprayed with lemon juice mixture.

Therefore, the IV, DV, CV, EG and CG of this experiment are as follows:

  1. The independent variable (IV) is the lemon juice mixture
  2. The dependent variable (DV) is the appearance of the green slime on the shower
  3. The control variable (CV) are time taken to spray, the amount of spray
  4. The experimental group (EG) is the side of the shower sprayed with lemon juice mixture
  5. The control group (CG) is the side of the shower sprayed with water.

Learn more: brainly.com/question/17498238?referrer=searchResults

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