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AlladinOne [14]
2 years ago
14

Explain imprinting in relation to Lorenz's geese. What would most likely have happened if the first thing the goslings saw was a

dog?
Physics
1 answer:
iragen [17]2 years ago
7 0

If the first thing that the goslings saw was a dog, they would have followed the dog as a mother.

Imprinting refers to the process of training an animal to bond with anything it sees after birth even if it is not its real mother. Lorenz first achieved imprinting in 1935 using geese which followed him as their mother shortly after they were born.

If the geese were exposed to a dog, they could also have seen the dog as their mother and followed it accordingly shortly after birth.

Learn more: brainly.com/question/11401513

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garik1379 [7]

Answer:

The parent cell

Explanation:

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3 years ago
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One number is said to be an "order of magnitude" larger than another number if
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Im thinking D. It is 100 times larger

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3 years ago
A 63-kg hiker is climbing the 828-m-tall Burj Khalifa in Dubai. If the efficiency of converting the energy content of the bars i
konstantin123 [22]

Answer:

T_f=5854.76 °C

Explanation:

Given:

mass of hiker, m= 63 kg

height to be climbed, h= 828 m

energy produced by an energy bar, E= 1.10\times 10^6 J

heat capacity of the hiker, c=75.3 J.mol^{-1}.K^{-1}= 4.184 J.kg^{-1}.K^{-1}

initial body temperature of hiker, T_i=36.6 \degree C

<em>The efficiency of converting the energy content of the bars into the work of climbing is 25%, the remaining 75% of the energy released through metabolism is heat released to her body.</em>

We find the energy required for climbing 828 m height:

W=m.g.h

W=63\times 9.8\times 828

W= 511207.2 J

∵Hike eats 2 energy bars= 2\times 1.1\times 10^{6} J

Energy produced= 2.2\times 10^{6} J

Now, according to her efficiency:

Total energy required for producing the work of W= 511207.2 J which is required to climb the given height will be (say, E):

25\% of E= 511207.2

\Rightarrow E= 511207.2\times \frac{100}{25}

E=2044828.8 J

&

Amount of total energy (E) converted into heat(Q) is:

Q=2044828.8-511207.2\\Q=1533621.6J

As we know that:

heat, Q=m.c. (T_f-T_i).................(1)

where:

T_f is the final temperature

Putting respective values in the eq. (1)

1533621.6= 63\times 4.184\times (T_f-36.6)

(T_f-36.6)\approx 5818.16

T_f\approx 5854.76 °C

4 0
3 years ago
a0.155 kg arrow is shot from ground level, upward at 31.4 m/s. what is its kinetic energy (ke) when it is 30.0 m above the groun
swat32

The kinetic energy (KE) of a 0.155 kg arrow that is shot from ground level, upward at 31.4 m/s, when it is 30.0 m above the ground is 30.85 J

Assuming air friction is negligible,

a = - 9.8 m / s²

u = 31.4 m / s

s = 30 m

v² = u² + 2 a s

v² = 31.4² + ( 2 * - 9.8 * 30 )

v² = 985.96 - 588

v² = 397.96 m / s

KE = 1 / 2 m v²

KE = 1 / 2 * 0.155 * 397.96

KE = 0.0775 * 397.96

KE = 30.85 J

Therefore, the kinetic energy ( KE ) when it is 30.0 m above the ground is 30.85 J

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Fast breeder reactor technology definition
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