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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! :)

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The manipulation of natural sounds via the medium of magnetic tape is called
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3 years ago
A body is placed on a rough inclined plane. Why does the frictional force decrease with the increase of angle of inclination
kondor19780726 [428]
The frictional force is directly proportional to the force that is perpendicular on the surface.

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Now as the inclination of the surface increases, the gravitational force is no longer the perpendicular force of the body, its value decreases, which means only a part is used to generate frictional force. Consequently, frictional force decreases.

When the inclination reaches 90 degrees, the gravitational force does not act along the normal and accordingly, no friction force is generated.

7 0
3 years ago
Three charged particles are positioned in the xy plane: a 50-nC charge at y = 6 m on the y axis, a –80-nC charge at x = –4 m on
leva [86]

Answer:48 V

Explanation:

Given

Three charged particle with charge

q_1=50\ nC at y=6\ m

q_2=-80\ nC at x=-4\ m

q_3=70\ nC at y=-6\ m

Electric Potential is given by

V=\frac{kQ}{r}

Distance of q_1 from x=8\ m

d_1=\sqrt{6^2+8^2}

d_1=\sqrt{36+64}

d_1=10\ m

similarly d_2=8-(-4)

d_2=12\ m

d_3=\sqrt{(-6)^2+8^2}

d_3=\sqrt{36+64}

d_3=10\ m

Potential at x=8\ m is

V_{net}=\frac{kq_1}{d_1}+\frac{kq_2}{d_2}+\frac{kq_3}{d_3}

V_{net}=k[\frac{q_1}{d_1}+\frac{q_2}{d_2}+\frac{q_3}{d_3}]

V_{net}=9\times 10^9[\frac{50}{10}-\frac{80}{12}+\frac{70}{10}]\times 10^{-9}

V_{net}=9\times 5.33

V_{net}=47.97\approx 48\ V

5 0
3 years ago
An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ \frac{du}{dy}

so

= µ  \frac{v}{h}   ............1

put here value

= 1.75×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

and

area between air and puck is given by

Area = \frac{\pi }{4} d^{2}

area  =  \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v    

and now apply newton second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

t =  \frac{0.1 v * 0.03}{1.37*10^{-3} v}

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0
3 years ago
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