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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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HELP ASAP PLEASE SCIENCE
shutvik [7]

Answer:

I think it's claim 1 because the same mass would cause them both to either fall off or to stay there

3 0
3 years ago
Consider a turntable to be a circular disk of moment of inertia It rotating at a constant angular velocity ωi around an axis thr
Rom4ik [11]

Answer:

Note: Angular momentum is always conserved in a collision.

The initial angular momentum of the system is

L = ( It ) ( ωi )

where It = moment of inertia of the rotating circular disc,

ωi = angular velocity of the rotating circular disc

The final angular momentum is

L = ( It + Ir ) ( ωf )

where ωf is the final angular velocity of the system.

Since the two angular momenta are equal, we see that

( It ) ( ωi ) = ( It + Ir ) ( ωf )

so making ωf the subject of the formula

ωf = [ ( It ) / ( It + Ir ) ] ωi

Explanation:

7 0
3 years ago
An object with an initial speed of 4.0 meters per second accelerates uniformly at 2.0 meters per second squared in the direction
leva [86]

Answer:

6 m/s.

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 4 m/s

Acceleration (a) = 2 m/s²

Distance (s) = 5 m

Final velocity (v) =?

The final velocity of the object can obtained as shown below:

v² = u² + 2as

v² = 4² + (2 × 2 × 5)

v² = 16 + 20

v² = 36

Take the square root of both side.

v = √36

v = 6 m/s

Therefore, the final speed of the object is 6 m/s.

8 0
3 years ago
A small ball with mass 1.50 kg is mounted on one end of a rod 0.600 m long and of negligible mass. The system rotates in a horiz
Georgia [21]

Answer:

(A) 0.54 kg.m^{2}

(B)  0.0156 N

Explanation:

from the question you would notice that there are some missing details, using search engines you can find similar questions online here 'https://www.chegg.com/homework-help/questions-and-answers/small-ball-mass-120-kg-mounted-one-end-rod-0860-m-long-negligible-mass-system-rotates-hori-q7245149'

here is the complete question:

A small ball with mass 1.20 kg is mounted on one end of a rod 0.860 m long and of negligible mass. The system rotates in a horizontal circle about the other end of the rod at 5100 rev/min. (a) Calculate the rotational inertia of the system about the axis of rotation. (b) There is an air drag of 2.60 x 10^{-2} N on the ball, directed opposite its motion. What torque must be applied to the system to keep it rotating at constant speed?.

solution

mass of the ball (m) = 1.5 kg

length of the rod (L) = 0.6 m

angular velocity (ω) = 4900 rpm

air drag (F) =  2.60 x 10^{-2} N = 0.026 N

(take note that values from the original question are used, with the exception of the air drag which was not in the original question)

(A) because the rod is mass less, the rotational inertia of the system is the rotational inertia of the rod about the other end, hence rotational inertia =mL^{2} where m = mass of ball and L =  length of rod

= 1.5 x 0.6^{2} = 0.54 kg.m^{2}

(B) The torque that must be applied to keep the ball in motion at constant speed = FLsin90

= 0.026 x 0.6 x sin 90 = 0.0156 N

4 0
3 years ago
Leaving the distance between the 181 kg and the 712 kg masses fixed, at what distance from the 712 kg mass (other than infinitel
bezimeni [28]

Answer:

Explanation:

The force due to gravitation is equal to zero for each of the masses.

M1= 181kg

M2= 712kg

m = 72.6kg

The distance between M1 and M2 is said to be fixed , therefore no value should be given I.e it's a constant.

From the formula for gravitational force we have that

F = GMm/r^2

GmM1/(d-r)^2. = GmM2/r^2

where r is the distance between the 72.6 kg and 712kg

d is the distance between M1 and M2

Solving mathematically

r(√M1+√M2) = d√M2

r = d√M2/√M1 + √M2

d×26.68/ 13.45+26.68

d×26.68/40.13

r = 0.665d

4 0
4 years ago
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