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prohojiy [21]
2 years ago
14

A scientist counts the number of bacteria colonies growing in each of six lab dishes.

Physics
2 answers:
Vsevolod [243]2 years ago
5 0

Hi!

The correct answer is: C. It is the data value that appears the most often in the set.

<h3>Explanation</h3>

The mode of a set of values is the number which is the most frequently appearing, or most repeating value in the range. The numbers that show one time each are

A - 16 , C - 25 , D - 3

While 13 appears three times with the dishes labelled

B , E and F

So, the value that appears the most times is 13, and hence 13 is taken as the mode of this data set

<h3>Hope this helps!</h3>
Dafna1 [17]2 years ago
4 0

Answer:

the answer is c It is the data value that appears the most often in the set.

Explanation:

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Answer:

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3 years ago
A child has an ear canal that is 1.3 cm long. Assume the speed of sound is v = 344 m/s.
kap26 [50]

Answer:

The  frequencies are (f, f_1) =  (6615.4 \ Hz , 19846.2\ Hz)

Explanation:

From the question we are told that

  The  length of the ear canal is  l = 1.3 \ cm  =\frac{1.3}{100}  =  0.013 \ m

   The  speed of sound is assumed to be  v_s  =  344 \ m/s

Now  taking look at a typical  ear canal  we see that we assume it is  a  closed pipe

   Now the fundamental harmonics for the pipe(ear canal) is mathematically represented as

            f = \frac{v_s}{4 * l }

 substituting values  

          f = \frac{344}{4 * 0.013 }

         f = 6615.4 \ Hz

Also the the second harmonic for the pipe (ear canal) is mathematically represented as

        f_1 =  \frac{3v_s}{4 * l}

 substituting values  

       f_1 =  \frac{3 *  344}{4 * 0.013}

       f_1 =   19846.2 \ Hz

Given that sound would be loudest in the pipe at the frequency, it implies that the child  will have an increased audible sensitivity at this  frequencies

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3 years ago
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Which change will always result in an increase in the gravitational force between 2 objects
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3 years ago
Read 2 more answers
The spectral distribution of the radiation emitted by a diffuse surface may be approximated as follows. HW 2 Q2 (a) What is the
Ronch [10]

Answer:

a)<em> 2000 W/m²  </em><em>; </em>b) 636.94 W/m<em>².sr ; </em><em>c) </em>0.5

Explanation:

a)

The formula for calculation of total emissive power is:

Total emissive power = E = \int\limits^\alpha_0 E'<em>λdλ</em>

<em>                                    </em>= \int\limits^a_0(0)d<em>λ + </em>\int\limits^b_a(100)d<em>λ + </em>\int\limits^c_b(200)d<em>λ + </em>\int\limits^d_c(100)d<em>λ </em>\int\limits^e_d(0)d<em>λ</em>

<em>where a = 5; b = 10; c = 15; d = 20; e = 25</em>

<em>                                    = 0 +100(10-5) + 200(15-10) +100(20-15) + 0</em>

<em>                                    = 2000 W/m²</em>

b)

The formula for total intensity of radiation is:

I_{e} = E/π = 200/3.14 = 636.94 W/m<em>².sr  </em>

<em>c)</em>

Fo submissive power leaving the surface in range π/4 ≤θ≤π/2

[E(π/4 ≤θ≤π/2)]/E = \int\limits^f_0\int\limits^g_0\int\limits^i_h Icosθsinθ dθdΦdλ

where f = infinity, g=2π, h=π/4, i=π/2

By simplifying, we get

                           = (-1/2)[cos(2π/2)-cos(2π/2)]

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                           =0.5

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