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Lisa [10]
3 years ago
14

What are some characteristics of an Effective Early Childhood teacher ​

Physics
2 answers:
stellarik [79]3 years ago
8 0

Answer:

To be a teacher in early childhood education, you need:

• Have training in early childhood education.

• Have the ability to encourage, motivate children.

• Patience and tolerance.

• Enthusiasm and dedication.

• Listen and observe.

• Have the ability to express themselves clearly, both orally and in writing.

• Know how to maintain discipline.

• Organize and plan.

Explanation:

Early childhood education teachers help the physical, affective, communicative, social and cognitive development of all children between the ages of zero and five. They are professionals who actively work on the development of the child at this stage of life. Its activity is divided according to: children from zero to three years of age in nurseries; and from three to five years of age in ordinary schools, early childhood education cycle.

nata0808 [166]3 years ago
6 0

Answer:

Enthusiasm, patience, and the respect of differences.

Explanation:

It takes courage and time for an effective early childhood teacher to understand and appreciate what she has with the kids and to understand them and help kids with creativity and to make them feel awesome.

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A satellite that weighs 4900 N on the launchpad travels around the earth's equator in a circular orbit with a period of 1.667 h.
charle [14.2K]

Answer:

(a)F= 3.83 * 10^3 N

(b)Altitude=8.20 * 10^5 m

Explanation:

On the launchpad weight = gravitational force between earth and satellite.

W = GMm/R²

where R is the earth radius.

Re-arranging:

WR² / GM = m

m = 4900 * (6.3 * 10^6)² / (6.67 * 10^-11 * 5.97 * 10^24) = 488 kg

The centripetal force (Fc) needed to keep the satellite moving in a circular orbit of radius (r) is:

Fc = mω²r

where ω is the angular velocity in radians/second. The satellite completes 1 revolution, which is 2π radians, in 1.667 hours.

ω = 2π / (1.667 * 60 * 60) = 1.05 * 10^-3 rad/s

When the satellite is in orbit at a distance (r) from the CENTRE of the earth, Fc is provided by the gravitational force  between the earth and the satellite:

Fc = GMm/r²

mω²r = GMm / r²

ω²r = GM / r²

r³ = GM/ω² = (6.67 * 10^-11 * 5.97 * 10^24) / (1.05 * 10^-3)²  

r³ = 3.612 * 10^20

r = 7.12 * 10^6 m

(a) F = GMm/r²  

F=(6.67 * 10^-11 * 5.97 * 10^24 * 488) / (7.12 * 10^6 )²

F= 3.83 * 10^3 N

(b) Altitude = r - R = (7.12 * 10^6) - (6.3 * 10^6) = 8.20 * 10^5 m

4 0
3 years ago
Which image depict electrons moving so that each element has a stable noble-gas electron configuration ?
vitfil [10]
The fact I can get is c and d
5 0
4 years ago
A crane raises a crate with a mass of 150 kg to a height of 20 m. Given that
Virty [35]

Answer:

\boxed {\boxed {\sf 29,400 \ Joules}}

Explanation:

Gravitational potential energy is the energy an object possesses due to its position. It is the product of mass, height, and acceleration due to gravity.

E_P= m \times g \times h

The object has a mass of 150 kilograms and is raised to a height of 20 meters. Since this is on Earth, the acceleration due to gravity is 9.8 meters per square second.

  • m= 150 kg
  • g= 9.8 m/s²
  • h= 20 m

Substitute the values into the formula.

E_p= 150 \ kg \times 9.8 \ m/s^2 \times 20 \ m

Multiply the three numbers and their units together.

E_p=1470 \ kg*m/s^2 \times 20 m

E_p=29400 \ kg*m^2/s^2

Convert the units.

1 kilogram meter square per second squared (1 kg *m²/s²) is equal to 1 Joule (J). Our answer of 29,400 kg*m²/s² is equal to 29,400 Joules.

E_p= 29,400 \ J

The crate has <u>29,400 Joules</u> of potential energy.

7 0
3 years ago
A man started at his house and went to the post office, then to the pet store, and finally home again. What is the average veloc
docker41 [41]

Answer:

0

Explanation:

Defining velocity :

Velocity is a vector which is the ratio of a person's total displacement with time.

Displacement, in simple terms refers to the distance between an individual's initial position to his final position.

Man's initial position = House

After all his navigation and points covered irrespective of the distance ;

Final position = House

Hence, we can conclude that the man's Displacement is ;

Final position - Initial position = 0

Hence,

Velocity = Displacement / time taken

Velocity = 0 / 3

Velocity = 0

6 0
3 years ago
Jada was walking home for 30 mins. How fast was she walking, if her house is 4 km away fron school?
earnstyle [38]

Answer:

8 km an hour (about 5mph)

Explanation:

It depends on what you are looking for because there is no unit specified, but if you want kilometers an hour then all you have to do is multiply both sides by 2.

30 x 2 = 60 (minutes)

4 x 2 = 8 (kilometers)

Hope this helped at least a little bit!

3 0
3 years ago
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