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Ivahew [28]
3 years ago
10

Natalie and Will are discussing socialization. Natalie says that socialization occurs when an animal becomes accustomed to the p

eople in the household. Will says that socialization is easily attained if the animal is first exposed to humans after 12 weeks of age. Who is correct?
A. Only Natalie
B. Only Will<~~~My answer and its wrong :(
C. Neither Natalie nor Will
D. Both Natalie and Will
Physics
2 answers:
skad [1K]3 years ago
7 0
The correct answer for this question is this one: "C. Neither Natalie nor Will." Natalie and Will are discussing socialization. Natalie says that socialization occurs when an animal becomes accustomed to the people in the household. <span>Will says that socialization is easily attained if the animal is first exposed to humans after 12 weeks of age.</span>
RSB [31]3 years ago
4 0

Answer:

A. only Natalie

Explanation:

The answer above is wrong. The correct answer is A "only Natalie" because the socialization process, while best done during the 2-12 weeks, it does not happen only in that period. As long as the animal gets accustomed to living in the household, it is being socialized

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True or false Carbon in the form of carbon dioxide is needed for both processes of photosynthesis and cellular respiration True
Anna007 [38]

Yes carbon dioxide is needed for photosynthesis while cellular respiration needs oxygen and dispurses carbon dioxide

5 0
3 years ago
In 1976, the SR-71A, flying at 20 km altitude (T = –56 0C), set the official jet-powered aircraft speed record of 3530 km/hr (21
Lapatulllka [165]

To solve this problem we will apply the concepts related to the calculation of the speed of sound, the calculation of the Mach number and finally the calculation of the temperature at the front stagnation point. We will calculate the speed in international units as well as the temperature. With these values we will calculate the speed of the sound and the number of Mach. Finally we will calculate the temperature at the front stagnation point.

The altitude is,

z = 20km

And the velocity can be written as,

V = 3530km/h (\frac{1000m}{1km})(\frac{1h}{3600s})

V = 980.55m/s

From the properties of standard atmosphere at altitude z = 20km temperature is

T = 216.66K

k = 1.4

R = 287 J/kg

Velocity of sound at this altitude is

a = \sqrt{kRT}

a = \sqrt{(1.4)(287)(216.66)}

a = 295.049m/s

Then the Mach number

Ma = \frac{V}{a}

Ma = \frac{980.55}{296.049}

Ma = 3.312

So front stagnation temperature

T_0 = T(1+\frac{k-1}{2}Ma^2)

T_0 = (216.66)(1+\frac{1.4-1}{2}*3.312^2)

T_0 = 689.87K

Therefore the temperature at its front stagnation point is 689.87K

6 0
4 years ago
When light goes from one material into another material having a HIGHER index of refractionA) its speed decreases but its wavele
Mnenie [13.5K]

Answer:

E) its speed and wavelength decrease, but its frequency stays the same

Explanation:

First of all, the frequency of a light wave does not depend on the medium, while wavelength and speed do. Therefore, the frequency remains costant.

In particular, the speed of light in a medium is given by:

v=\frac{c}{n}

where c is the speed of light in a vacuum and n is the index of refraction. From the formula, we see that v and n are inversely proportional: so, when the light moves into a material with higher index of refraction, its speed decreases.

Moreover, speed is related to wavelength by

v=\lambda f

where \lambda is the wavelength and f is the frequency. Since the two quantities are directly proportional, this means that since the speed decreases, the wavelength decreases as well.

So, the correct choice is

E) its speed and wavelength decrease, but its frequency stays the same

5 0
3 years ago
Read 2 more answers
Contact lenses are designed to be gas permeable in order to allow oxygen to diffuse to the eye. If the edge of a contact lens is
guapka [62]

Answer: A

The diffusion time through the edge is double the diffusion time through the center.

Explanation:

Only the diffusion coefficient, thickness, surface area, distance affects the time it takes a gas such as oxygen to diffuse a given in a medium. The Diffusion time increases with the square of diffusion distance  Also, the Increased surface area increases the rate of diffusion or the diffusion time, whereas a thicker membrane reduces the rate of diffusion or time it takes the gas to be permeable. Also, The greater the distance that a substance must travel, the slower the rate of diffusion The diffusion coefficient determines the time it takes a gas to diffuse. Also the Diffusion time is inversely proportional to the diffusion coefficient.

Now, If the edge of a contact lens is double the thickness of the central portion of the lens, then it will take more time for the gas to diffuse through the thicker portion. since its edge is double the central position, then the diffusion time will also be doubled. So Option A is the best answer-

The diffusion time through the edge is double the diffusion time through the center.

6 0
4 years ago
How much work is required to lift a 10-newton weight from 4.0 meters to 40 meters above the surface of Earth?
kumpel [21]
<h3>Answer : 360J</h3>

<h3>Way to do : </h3>

s = 40m - 4m = 36m

W = F × s

= 10N × 36m = 360J

<h3>A bit of explanation : </h3>

W = Work (J)

F = Force / weight (N)

s = distance (m)

6 0
3 years ago
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