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allochka39001 [22]
3 years ago
13

a scientist divided a population of fruit flies into two containers, each with a different kind of food. then she bred the fruit

flies under these conditions for many generations. lastly, she put the fruit flies from the two containers together again and observed which flies mated. which question was the scientist most likely testing? a. do well-fed fruit flies make good pets? b. what kind of food do fruit flies prefer? c. how many species of fruit flies can mate with each other? d. will fruit flies bred under different conditions mate?
Physics
1 answer:
kati45 [8]3 years ago
4 0
<span>d. will fruit flies bred under different conditions mate? Let's look at the possible choices and see which of them make any sense given the experiment. a. do well-fed fruit flies make good pets? Seems kinda silly, but if this were the question being asked, I suspect the experiment would have some fruit flies that were well fed as well as some fruit flies that were starved and would then compare how those 2 populations of fruit flies interacted with people. But that wasn't done, so it's unlikely this is the question being asked. b. what kind of food do fruit flies prefer? Each population of fruit flies weren't given a choice as to available foods. So their preferences didn't come into play about what they were allowed to eat. So this question is also unlikely. c. how many species of fruit flies can mate with each other? The scientist started with a single population and divided it into two sub groups. There doesn't seem to be a large number of species of different fruit flies here, so this is also an unlikely question. d. will fruit flies bred under different conditions mate? The scientist started with a single population of fruit flies and divided them into two groups. Each group was allowed to breed for many generations with a different food for each group (e.g. Allowed to breed under different conditions). After they were both well established, the groups were merged together and observed which ones mated. It looks like this question is being answered. So this is the correct solution.</span>
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What is the magnitude of your displacement when you follow directions that tell you to walk 225 m in one direction, make a 90° t
Gekata [30.6K]

Start by facing East. Your first displacement is the vector

<em>d</em>₁ = (225 m) <em>i</em>

Turning 90º to the left makes you face North, and walking 350 m in this direction gives the second displacement,

<em>d</em>₂ = (350 m) <em>j</em>

Turning 30º to the right would have you making an angle of 60º North of East, so that walking 125 m gives the third displacement,

<em>d</em>₃ = (125 m) (cos(60º) <em>i</em> + sin(60º) <em>j</em> )

<em>d</em>₃ ≈ (62.5 m) <em>i</em> + (108.25 m) <em>j</em>

The net displacement is

<em>d</em> = <em>d</em>₁ + <em>d</em>₂ + <em>d</em>₃

<em>d</em> ≈ (287.5 m) <em>i</em> + (458.25 m) <em>j</em>

and its magnitude is

|| <em>d</em> || = √[ (287.5 m)² + (458.25 m)² ] ≈ 540.973 m ≈ 541 m

7 0
3 years ago
An internal explosion breaks an object, initially at rest, into two pieces: A and B. Piece A has 1.9 times the mass of piece B.
Helen [10]

Kinetic energy of pieces A and B are 2724 Joule and 5176 Joule respectively.

<h3>What is the relation between the masses of A and B?</h3>
  • Let mass of piece A = Ma

Mass of piece B = Mb

  • Velocities of pieces A and B are Va and Vb respectively.
  • As per conservation of momentum,

Ma×Va = Mb×Vb

  • Here, Ma=1.9Mb

So, 1.9Mb × Va = Mb×Vb

=> 1.9Va = Vb

<h3>What are the kinetic energy of piece A and B?</h3>
  • Expression of kinetic energy of piece A = 1/2 × Ma × Va²
  • Kinetic energy of piece B = 1/2 × Mb × Vb²
  • Total kinetic energy= 7900J

=>1/2 × Ma × Va² + 1/2 × Mb × Vb² = 7900

=> 1/2 × Ma × Va² + 1/2 × (Ma/1.9) × (1.9Va)² = 7900

=> 1/2 × Ma × Va² ×(1+1.9) = 7900 j

=> 1/2 × Ma × Va² = 7900/2.9 = 2724 Joule

  • Kinetic energy of piece B = 7900 - 2724 = 5176 Joule

Thus, we can conclude that the kinetic energy of piece A and B are 2724 Joule and 5176 Joule respectively.

Learn more about the kinetic energy here:

brainly.com/question/25959744

#SPJ1

5 0
2 years ago
If the PLATE SEPARATION of an isolated charged parallel-plate capacitor is doubled: A. the electric field is doubled
mash [69]

Explanation:

The electric field of an isolated charged parallel-plate capacitor is given by :

E=\dfrac{q}{A\epsilon_o}........(1)

Where

q is the electric charge

A is the area of cross section of parallel plate

It is clear from equation (1) that the electric field of a parallel plate capacitor is directly proportional to the charge on the plate and inversely proportional to the area of cross section of a plate.

So, the correct option is (E) i.e. "none of the above".

5 0
3 years ago
A metal rod has a moves with a constant velocity of 40 cm/s along two parallel metal rails through a magnetic field of 0.575 T.
love history [14]

Answer:

2.12/R mW

Explanation:

The electrical power, P generated by the rod is

P = B²L²v²/R where B = magnetic field = 0.575 T, L = length of metal rod = separation of metal rails = 20 cm = 0.2 m, v = velocity of metal rod = 40 cm/s = 0.4 m/s and R = resistance of rod = ?

So, the induced emf on the conductor is

E = BLv

= 0.575 T × 0.2 m × 0.4 m/s

= 0.046 V

= 46 mV

The electrical power, P generated by the rod is

P = B²L²v²/R

=  B²L²v²/R

So, P = (0.575 T)² × (0.2 m)² × (0.4 m/s)²

= 0.002116/R W

= 2.12/R mW

3 0
3 years ago
Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton
pishuonlain [190]

Hello!

Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton force is applied to it.

Data:

Hooke represented mathematically his theory with the equation:

F = K * Δx  

On what:

F (elastic force) = 2 N

K (elastic constant) = 4 N/cm

Δx (deformation or elongation of the elastic medium or distance from a spring) = ?

Solving:


F = K * \Delta{x}

2\:N = 4\:N/cm*\Delta{x}

4\:N/cm*\Delta{x} = 2\:N

\Delta{x} = \dfrac{2\:\diagup\!\!\!\!\!N}{4\:\diagup\!\!\!\!\!N/cm}

simplify by 2

\Delta{x} = \dfrac{2}{4}\frac{\div2}{\div2}

\boxed{\boxed{\Delta{x} = \dfrac{1}{2}\:cm}}\Longleftarrow(distance)\end{array}}\qquad\checkmark

Answer:

B.) 1/2 cm

_______________________

I Hope this helps, greetings ... Dexteright02! =)

7 0
3 years ago
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