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Nadusha1986 [10]
3 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]3 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]3 years ago
3 0
Fruit flies prefer mates adapted to the same food source.

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Is calculating the change of velocity the same as calculating acceleration? ​
koban [17]

Answer:

Yes! Thinking about it graphically a position vs time graph models meters per second in most cases, making every point on the line have the units m/s. If we want the find the slope we are finding the change between each point and those units would change to m/s/s or m/s^2 giving us the same units for acceleration. Simply put, slope of a velocity graph gives us acceleration.

Explanation:

4 0
2 years ago
What is the relationship of an object's acceleration, its mass and the force applied to it?
Anon25 [30]

Answer:

F = ma

Explanation:

Newtons second law

8 0
3 years ago
Hans Langseth's beard measured 5.33 m in 1927. Consider two charges, q1 = 2.42 nC and an unspecified charge, q2, are separated 5
schepotkina [342]

Answer:

-7.89 * 10^(-9) C

Explanation:

Parameters given:

q1 = 2.42 nC = 2.42 * 10^(-9) C

Distance between q1 and q2 = 5.33 m

q3 = 1.0 nC = 1 * 10^(-9) C

Distance between q1 and q3 = 1.9 m

Distance between q2 and q3 = 5.33 - 1.9 = 3.43 m

The net force acting on q3 is:

F = F(q1, q3) + F(q2, q3)

F = (k*q1*q3)/1.9² + (k*q2*q3)/3.43²

F = (9 * 10^(9) * 2.42 * 10^(-9) * 1 * 10^(-9))/3.61 + (9 * 10^(9) * q2 * 1 * 10^(-9))/11.7649

F = 6.033 * 10^(-9) + 0.765*q2

If the net force is zero:

0 = 6.033 * 10^(-9) + 0.765*q2

-0.765*q2 = 6.033 * 10^(-9)

=> q2 = -[6.033 * 10^(-9)]/0.765

q2 = -7.89 * 10^(-9) C

3 0
3 years ago
A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle
ra1l [238]

Answer:

a) t = 4.16 s

b) x = 141.51 m

Explanation:

Given

v = 21.5 m/s

x0 = 52.0 m

a = 6.0 m/s²

a) Motorcycle

x = v0*t + (a*t²/2)

x = 21.5t + (6*t²/2)

x = 21.5t + 3t²   <em>(I)</em>

Car

x = x0 + v0*t

x = 52 + 21.5t  <em>(II)</em>

<em />

then we can apply <em>I = II</em>

21.5t + 3t² = 52 + 21.5t

⇒ 3t² = 52

⇒ t = 4.16 s

b) We can use <em>I</em> or <em>II</em>, then

x = 52 + 21.5*(4.16)

⇒ x = 141.51 m

8 0
3 years ago
You are sitting on a merry-go-round of mass 200 kg and radius 2m that is at rest (not spinning). Your mass is 50 kg. Your friend
Bogdan [553]

Answer:

a.\tau=200J b.\alpha=0.44 \frac{rad}{s^2} c. \alpha=0.33\frac{rad}{s^2} d. The angular acceleration when sitting in the middle is larger.

Explanation:

a. The magnitude of the torque is given by \tau=rF\sin \theta, being r the radius, F the force aplied and \theta the angle between the vector force and the vector radius. Since \theta=90^{\circ}, \, \sin\theta=1 and so \tau=rF=2m100N=200Nm=200J.

b. Since the relation \tau=I\alpha hols, being I the moment of inertia, the angular acceleration can be calculated by \alpha=\frac{\tau}{I}. Since we have already calculated the torque, all left is calculate the moment of inertia. The moment of inertia of a solid disk rotating about an axis that passes through its center is I=\frac{1}{2}Mr^2, being M the mass of the disk. If we assume that a person has a punctual mass, the moment of inertia of a person would be given by I_p=m_pr_p^{2}, being m_p the mass of the person and r_p^{2} the distance from the person to the center. Given all of this, we have

\alpha=\frac{\tau}{I}=\frac{\tau}{I_{disk}+I_{person}}=\frac{Fr}{\frac{1}{2}Mr^2+m_pr_p^{2}}=\frac{200Nm}{\frac{1}{2}200kg*4m^2+50kg*1m^2}=\frac{200\frac{kgm^2}{s^2}}{450Nm^2}\approx 0.44\frac{rad}{s^2}.

c. Similar equation to b, but changing r_p=2m, so

alpha=\dfrac{200\frac{kgm^2}{s^2}}{\frac{1}{2}200*4kg\,m^2+50*4 kg\,m^2}=\dfrac{200}{600}\dfrac{1}{s^2}\approx 0.33 \frac{rad}{s^2}.

d. The angular acceleration when sitting in the middle is larger because the moment of inertia of the person is smaller, meaning that the person has less inertia to rotate.

5 0
3 years ago
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