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Nadusha1986 [10]
4 years ago
10

A scientist bred fruit flies in two separate containers with different food sources for many generations. When she put the fruit

flies from the two containers together again, she observed that the flies preferred mates raised on the same food. Which conclusion can the scientist draw?
Physics
2 answers:
Nadya [2.5K]4 years ago
5 0

Answer:

D. Fruit flies prefer mates adapted to the same food source.

Explanation:

According to a different source, these are the options included with this question:

A. Fruit flies mate randomly regardless of how they were bred.

B. Fruit flies can adapt to eating different food sources.

C. Fruit flies prefer the container where they were bred.

D. Fruit flies prefer mates adapted to the same food source.

The conclusion that the researcher can reach is that fruit flies prefer mates that are adapted to the same food source. In this example, we learn that when flies are raised separately, and have different food sources, they have been shown to prefer mates that were raised on the same food. This can successfully be extrapolated to conclude that fuit flies prefer mates that are adapted to the same food source.

Nookie1986 [14]4 years ago
3 0
Fruit flies prefer mates adapted to the same food source.

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A: No because it is nor changing speed or direction

B: Yes because it changes direction even though the speed is constant

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THE ELECTROMAGNET SPECTRUM ONLY HAS WHICH SHAPE OF WAVE?
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A Michelson interferometer uses light from a sodium lamp. Sodium atoms emit light having wavelengths 589.0 nm and 589.6 nm. The
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The distance mirror M2 must be moved so that one wavelength has produced one more new maxima than the other wavelength is;

<u><em>L = 57.88 mm</em></u>

<u><em /></u>

We are given;

Wavelength 1; λ₁ = 589 nm = 589 × 10⁻⁹ m

Wavelength 2; λ₂ = 589.6 nm = 589.6 × 10⁻⁹ m

We are told that L₁ = L₂. Thus, we will adopt L.

Formula for the number of bright fringe shift is;

m = 2L/λ

Thus;

For Wavelength 1;

m₁ = 2L/(589 × 10⁻⁹)

For wavelength 2;

m₂ = 2L/(589.6)

Now, we are told that one wavelength must have produced one more new maxima than the other wavelength. Thus;

m₁ - m₂ = 2

Plugging in the values of m₁ and m₂ gives;

(2L/589) - (2L/589.6) = 2

divide through by 2 to get;

L[(1/589) - (1/589.6)] = 1

L(1.728 × 10⁻⁶) = 1

L = 1/(1.728 × 10⁻⁶)

L = 578790.67 nm

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8 0
2 years ago
At what height h above the ground does the projectile have a speed of 0.5v?
maw [93]

Answer:

h=\dfrac{3v^2}{8g}

Explanation:

It is given that,

Speed of the projectile is 0.5 v. Let h is the height above the ground. Using the first equation of motion to find it.

v=u+at

v=u-gt

Initial speed of the projectile is v and final speed is 0.5 v.

0.5v=v-gt

t=\dfrac{v}{2g}

g is the acceleration due to gravity

Let h is the height above the ground. Using the second equation of motion as :

h=vt-\dfrac{1}{2}gt^2

h=v\dfrac{v}{2g}-\dfrac{1}{2}g(\dfrac{v}{2g})^2

h=\dfrac{3v^2}{8g}

So, the height of the projectile above the ground is \dfrac{3v^2}{8g}. Hence, this is the required solution.

6 0
3 years ago
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