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lina2011 [118]
3 years ago
13

A rancher wants to investigate the use of artificial selection in his cattle herd. Which of these options is an example of artif

icial selection?
A) Breed a cow that produces large quantities of milk with a bull that comes from a line of cattle with long lifespans.

B) Trace the family tree of a chicken that lays many eggs back through several generations.

C) Separate a population of rabbits and monitor any trait changes over the next few generations.

D) Choose one individual in a population of sheep to favor with food and shelter and see if it mates more.
Physics
1 answer:
Viefleur [7K]3 years ago
4 0

Answer: The correct answer is A. Breed a cow that produces large quantities of milk with a bull that comes from a line of cattle with long lifespans.

Explanation: This answer makes the most since and it shows the use of artificial selection. Artificial selection is when humans breed selectively to develop certain traits they are looking for or want in an animal. (I took the test and this was the correct answer.)

Have a nice day! :)

You might be interested in
Planet 1 orbits Star 1 and Planet 2 orbits Star 2 in circular orbits of the same radius. However, the orbital period of Planet 1
hichkok12 [17]

Answer:

The mass of Star 2 is Greater than the mass of Start 1. (This, if we suppose the masses of the planets are much smaller than the masses of the stars)

Explanation:

First of all, let's draw a free body diagram of a planet orbiting a star. (See attached picture).

From the free body diagram we can build an equation with the sum of forces between the start and the planet.

\sum F=ma

We know that the force between two bodies due to gravity is given by the following equation:

F_{g} = G\frac{m_{1}m_{2}}{r^{2}}

in this case we will call:

M= mass of the star

m= mass of the planet

r = distance between the star and the planet

G= constant of gravitation.

so:

F_{g} =G\frac{Mm}{r^{2}}

Also, if the planet describes a circular orbit, the centripetal force is given by the following equation:

F_{c}=ma_{c}

where the centripetal acceleration is given by:

a_{c}=\omega ^{2}r

where

\omega = \frac{2\pi}{T}

Where T is the period, and \omega is the angular speed of the planet, so:

a_{c} = ( \frac{2\pi}{T})^{2}r

or:

a_{c}=\frac{4\pi^{2}r}{T^{2}}

so:

F_{c}=m(\frac{4\pi^{2}r}{T^{2}})

so now we can do the sum of forces:

\sum F=ma

F_{g}=ma_{c}

G\frac{Mm}{r^{2}}=m(\frac{4\pi^{2}r}{T^{2}})

in this case we can get rid of the mass of the planet, so we get:

G\frac{M}{r^{2}}=(\frac{4\pi^{2}r}{T^{2}})

we can now solve this for T^{2} so we get:

T^{2} = \frac{4\pi ^{2}r^{3}}{GM}

We could take the square root to both sides of the equation but that would not be necessary. Now, the problem tells us that the period of planet 1 is longer than the period of planet 2, so we can build the following inequality:

T_{1}^{2}>T_{2}^{2}

So let's see what's going on there, we'll call:

M_{1}= mass of Star 1

M_{2}= mass of Star 2

So:

\frac{4\pi^{2}r^{3}}{GM_{1}}>\frac{4\pi^{2}r^{3}}{GM_{2}}

we can get rid of all the constants so we end up with:

\frac{1}{M_{1}}>\frac{1}{M_{2}}

and let's flip the inequality, so we get:

M_{2}>M_{1}

This means that for the period of planet 1 to be longer than the period of planet 2, we need the mass of star 2 to be greater than the mass of star 1. This makes sense because the greater the mass of the star is, the greater the force it applies on the planet is. The greater the force, the faster the planet should go so it stays in orbit. The faster the planet moves, the smaller the period is. In this case, planet 2 is moving faster, therefore it's period is shorter.

6 0
3 years ago
a 2000 kg truck is moving eastward at 25 m/s. it collides inelastically with a 1500 kg truck traveling southward at 30 m/s. they
Otrada [13]

According to the given statement Final velocity when they stick together is 8.735i^ + 11.25j^​

<h3>What is collision and momentum?</h3>

The unit of momentum is kg ms -1. Momentum is a vector parameter that is influenced by the object's direction. During collisions involving objects, momentum is a relevant concept. The final velocity before a collision between two objects equals the total motion after the impact (in the absence of external forces).

<h3>Briefing:</h3>

From conservation of momentum

Initial momentum = final momentum

m u +M U =(m+M) V

2000×25 i^ +1500×30 j^​ =(2000+1500) V

V = 8.735i^ + 11.25j^​

Final velocity when they stick together is 8.735i^ + 11.25j^​

To know more about Collide visit:

brainly.com/question/27993473

#SPJ4

The complete question is -

A 2000 kg truck is moving eastward at 25 m/s. it collides inelastically with a 1500 kg truck traveling southward at 30 m/s. they collide at the intersection. Find the direction and magnitude of velocity of the wreckage after the collision, assuming the vehicles stick together after the collision.

6 0
1 year ago
A large flat block of mass 7 m rests on a level, smooth tabletop (µk ≈ 0). On top of it is a much smaller block of mass m, with
mariarad [96]

Answer:

3. µs g /7

Explanation:

The largest Force appear when the maximal friction Force is required.

Second Newton law for the small block:

F_friction=u_s*N=u_s*(mg)

F-F_friction=ma

F-u_s*(mg)=ma

Second Newton law for the Big Block:

F_friction=7ma

u_s*(mg)=7ma

a=u_s*g/7

8 0
3 years ago
Prompt
Mamont248 [21]

Answer:

1.bandages

2.cottons

3.scissor

4.medicines

6 0
3 years ago
A clock on a moving train runs __________ an identical clock at rest.
katovenus [111]

It depends who is watching it.

-- A passenger on the train says that his clock is normal, but the clock at the station he just passed is running slow.

-- A ticket agent at the station says that his clock is normal, but the clock on the train that just rolled through is running slow.

-- They're both correct, but ...

-- The difference is too small for either of them to notice, unless the train is going faster than about half the speed of light.

-- It isn't.

4 0
4 years ago
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