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fgiga [73]
3 years ago
7

If a total of 1250 offspring were obtained from your cross, determine the number of offspring that you would expect to obtain th

at represent a single crossover event between pr and vg if interference does not occur.
Physics
1 answer:
irakobra [83]3 years ago
3 0

Answer:

106 single crossover events

Explanation:

Total number of offspring = 1250

<u>Determine the number of offspring that would be expected </u>

( Rf between the genes vg and Pr ) * ( Total number of progeny )

  =   0.125 * 1250 ≈ 156 ( value for single and double crossovers )

next :

( value of the Rf of double crossovers ) * ( Total number of progeny )

= (0.322 * 0.125) = 0.04025  * 1250 = 50

Hence the number of offspring expected to obtain/represent a single crossover event between pr and vg

=  156 - 50 =  106

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The answer is λ₂ = 6.48 cm or  6.52 cm.

The out-of-tune guitar may have a wavelength between "6.48 cm" and "6.52 cm."

fb = |f2 − f1|

f₁ = 343/0.064

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f₂ = 5276.9 Hz ± 17 Hz

f₂ = 5293.9 Hz or 5259.9 Hz

Now, calculating the possible wavelengths:

λ = 343/ 5259.9  or 343/ 5293.9

λ₂ = 6.48 cm or 6.52 cm

<h3>Why is beat frequency important?</h3>

When two waves with almost identical frequencies traveling in the same direction collide at a certain location, beats are produced. The opposing beneficial and harmful disruption causes the sound to alternatively be loud and weak whenever two sound waves with different frequencies reach your ear. This is referred to as beating.

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fb = |f2 − f1|

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f' = \left (\dfrac{v + v_0}{v - v_s} \right) \times f

Where;

f' = The frequency the observer hears

f = Actual frequency of the wave

v = The velocity of the sound wave

v_o = The velocity of the observer

v_s = The velocity of the source

Where the observer is stationary, we have;

(i) When the source is moving in the direction of the observer

f' = \left (\dfrac{v }{v - v_s} \right) \times f

(ii) When the source is receding from the observer, we have;

f' = \left (\dfrac{v }{v + v_s} \right) \times f

Therefore;

(a) A person left behind on the platform

For a person left behind on the platform, we have that the radio source is receding, therefore, we have;

f' = \left (\dfrac{v }{v + v_s} \right) \times f

(1) Given that (v + v_s) > v, therefore, v < (v + v_s), f' < f, the frequency heard by the person left on the platform, f', is smaller (lower) than the frequency produced by the radio

(2) The frequency is not constant as the speed of the source is increasing while it under the acceleration due to gravity

(3) During the fall, the speed of the source continuously increases under the effect of gravitational attraction and therefore the frequency heard by the person on the platform becomes progressively smaller

(b) A person down below floating on a rubber raft

For the the person down below on the rubber raft, the radio source is advancing

Therefore, the radio source is moving towards the person at rest down on the rubber raft, therefore, we have;

f' = \left (\dfrac{v }{v - v_s} \right) \times f

(1) Given that (v - v_s) < v, therefore, f' > f, the frequency heard by the person down below floating on the rubber raft, f', is greater (higher) than the frequency produced by the radio

(2) The frequency is not constant as the speed of the source is increasing while it under the acceleration due to gravity

(3) During the fall, the speed of the source continuously increases under the effect of gravitational attraction and therefore the frequency heard by the person on the platform becomes progressively greater (higher)

Explanation:

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