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svetoff [14.1K]
2 years ago
13

The velocity of a given wave remains the same while the wavelength increases. What happens to the frequency of the wave?

Physics
1 answer:
Degger [83]2 years ago
5 0

Answer:

B. It decreases proportionally to the wavelength

Explanation:

The relationship between wavelength, velocity, and frequency of a wave is given as;

f = \frac{v}{\lambda}

where;

f is the frequency of the wave

v is the velocity of the wave

λ is the wavelength

From the equation above, the frequency of a wave is inversely proportional to its wavelength. As the wavelength increases, the frequency decreases and vice versa.

Therefore, the correct option is "B"

B. It decreases proportionally to the wavelength

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An atom with 4 protons, 5 neutrons, and 4 electrons has an atomic mass of _____ amu. (Enter a whole number.)
Oksi-84 [34.3K]
Atomic mass = number of protons + number of neutrons = 4+5 = 9 amu
6 0
3 years ago
A car traveling initially at 9.49 m/s accelerates at the rate of 0.988 m/s^2 for 3.05s. What is it’s velocity at the end of the
weeeeeb [17]

Answer:

12.50 m/s

Explanation:

Vi = 9.49 m/s

a = 0.988 m/s²

t = 3.05 s

Vf = ?

Vf = Vi + at

Vf = 9.49 + (0.988)(3.05)

Vf = 12.50 m/s

6 0
3 years ago
Read 2 more answers
All Physicsts over here plz help in these questions!!!!!!!!!
mrs_skeptik [129]
Hello!

First one we can use that PE=mgh so we have

4.37*10^5J/(9.12*10^3kg*9.80m/s^2)= 4.89m

Second one we can use Newton’s Second Law
F=ma and in this case F=mg so we have

g= 3.28*10^-2N/6*10^-3kg = 5.47m/s^2

Hope this helps. Any questions please ask. Thank you.
6 0
3 years ago
Read 2 more answers
3. A certain wire, 3 m long, stretches by 1.2 mm when under tension of 200 N. By how much does
nikitadnepr [17]

Answer:

The extension of the second wire is   e_2 = 0.0024 \  m =  2.4 mm

Explanation:

From the question we are told that

    The length of the wire is L  = 3 \ m

     The elongation of the wire is  e =  1.2mm =  \frac{1.2}{1000} =  0.0012 m

        The tension is F  =  200 \ N

       The length of the second wire is  L_2   =  6 \ m

     

Generally the Young's modulus(Y) of this material is  

        Y  = \frac{stress}{strain }

Where stress =  \frac{F}{A}

    Where A is the area which is evaluated as  

           A = \pi r^2

  and   strain = \frac{extention}{length} =  \frac{e}{L}

   So

        Y  = \frac{\frac{F}{\pi r^2 } }{ \frac{e}{L}  }

Since the wire are of the same material Young's modulus(Y)  is constant

So we have  

              \frac{F * L }{r^2 e}  =  \pi * Y = constant

              F * L   =  constant   * r^2 e

Now the ration between the first and the second wire is

         \frac{F_1}{F_2}  * \frac{L_1}{L_2} =  \frac{r*2_1}{r^2}  *  \frac{e_1}{e_2}

Since tension , radius are constant

   We have

           \frac{L_1}{L_2} =   \frac{e_1}{e_2}

substituting values

          \frac{3}{6} =   \frac{0.0012}{e_2}

          0.5 e_2 =  0.0012

         e_2 = \frac{ 0.0012  }{0.5}

          e_2 = 0.0024 \  m =  2.4 mm

3 0
3 years ago
How can a scientist ensure that his or her data are reliable?
laiz [17]
If the scientist repeats the experiment over and over and gets the same results. Also if the scientist peer reviews the experiment to make sure there is no bias in his or her results.

7 0
3 years ago
Read 2 more answers
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