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katrin2010 [14]
2 years ago
12

Should young children be treated as “little adults”? Based on what you have learned about development, do you think that is reas

onable? Why or why not?
Physics
2 answers:
belka [17]2 years ago
7 0

Since I’m not sure what you have been learning about child development I will just answer with my personal opinion. I hope my answer can be of help.



I personally think children definitely should not be treated as young adults. It is important for young children to experience that period of innocence and immaturity. Without this they will be deprived from having other relationships with children their age because they will not get along with them since they are more mature than others. This could cause them to ultimately distance themselves from others and have poor social skills. Children can not be treated as a adults after all they just came into this world! Treating a child like this could leave them depressed or mad when they get older. This is because they can never look back and laugh at their own silly but funny mistakes. Or remember having childish fun with friends or family. This may not be important but it will severely affect the child when he/she becomes older. Young children also go through important stages of their life. If we are to treat our children like they are adults they will not go through these stages and will not develop practical skills they will need in life. This can also be very dangerous because this could cause a child to detach themselves from their parents. It is EXTREMELY important for kids to be able to attach themselves to their parent and start to learn or mimic their behavior. If you treat them as if they are an adult they could detach from you because they feel a lack of love or sensitivity from you.
saul85 [17]2 years ago
6 0

Answer:

maybe if you posted a picture of the story i can help you

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A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 24.0° below the horizontal. The negligent
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Answer:

(a): The car's relative position to the base of the cliff is x= 32.52m.

(b): The lenght of the car in the ir is tfall= 1.78 sec.

Explanation:

Vo= 0

V= ?

d= 50m

h= 30m

a= 4 m/s²

t= √(2*d/a)

t= 5 sec

V= a*t

V= 20 m/s

Vx= V * cos(24º)

Vx= 18.27 m/s

Vy= V* sin(24º)

Vy= 8.13 m/s

h= Vy*t + g*t²/2

clearing t:

tfall= 1.78 sec (b)

x= Vx * tfall

x= 32.52 m (a)

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3 years ago
1. How many seconds in 1 year?
Nezavi [6.7K]

Answer:

3.154e+7

Explanation:

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2 years ago
In this problem, you will calculate the location of the center of mass for the Earth-Moon system, and then you will calculate th
Radda [10]

Answer:

a) Option D is correct.

The center of mass between the Eartg and the moon is inside the Earth.

Explanation:

Given,

Mass of the moon = (7.35×10²²) kg

Mass of the Earth = (6.00×10²⁴) kg

Mass of the Sun = (2.00×10³⁰) kg

Distance between the Earth and the moon = (3.80×10⁵) km

Distance between the Earth and the Sun = (1.50×10⁸) km

With the assumption that all.of the bodies being considered are on the same straight line on the x-axis,

Note that Centre of mass is given as

C.M = (Σmx)/(Σm)

For the Earth-moon system, let the earth be x=0, then the moon is at x = (3.80 × 10 5) km away.

C.M = (Σmx)/(Σm)

Σmx = (6.00×10²⁴)) × (0) + (7.35×10²²) × (3.80×10⁵) = (2.793 × 10²⁸) kg.km

Σm = (6.00×10²⁴) + (7.35×10²²) = (6.0735 × 10²⁴) kg

CM = (2.793 × 10²⁸) ÷ (6.0735 × 10²⁴)

CM = (4.60 × 10³) km = 4600 km

This means the centre of mass is 4600 km from the Earth.

The Earth's radius = 6378 km

Hence, the centre if mass is inside the Earth.

Hope this Helps!!!

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3 years ago
A solid circular disk has a mass of 1.2 kg and a radius of 0.16m. Each of three identical thin rods has a mass of 0.16kg. The ro
Juli2301 [7.4K]

Answer:

0.027648 kgm²

Explanation:

M = Mass of disc = 1.2 kg

r = Radius of disc = 0.16 m

m = Mass of rod = 0.16 kg

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Moment of inertia of disk is given by

I_1=\dfrac{1}{2}Mr^2\\\Rightarrow I_1=\dfrac{1}{2}1.2\times 0.16^2\\\Rightarrow I_1=0.01536\ kgm^2

Moment of inertia of the three rods

I_2=3mr^2\\\Rightarrow I_2=3\times 0.16\times 0.16^2\\\Rightarrow I_2=0.012288\ kgm^2

The total moment of inertia is given by

I=I_1+I_2=0.01536+0.012288\\\Rightarrow I=0.027648\ kgm^2

The moment of inertia of the stool with respect to an axis that is perpendicular to the plane of the disk at its center is 0.027648 kgm²

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