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Firdavs [7]
3 years ago
8

Why do we teach young children through song, rhyme, and rhythm? ?

Physics
2 answers:
Ronch [10]3 years ago
6 0

Answer:

gym and gym work 6.30 and I have been a very good friend a

german3 years ago
6 0

Answer:

Becoming aware of rhyming sounds boosts brain activity and a child's early literacy ability. Adding singsong rhyming words to requests for attention is an effective way for teachers to get toddlers to listen to what they say. Rhymes and rhythms add zest and humor and increase toddler cooperation in the classroom.

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Example 4:
earnstyle [38]

Answer:

Given speed of light in diamond =1.24*10m/s

speed of light (c) =3*10^8m/s

refractive index of diamond =3*10^8/1.24*10=2.41*10^7m/s

3 0
3 years ago
What is the magnitude of velocity for a 2,000 kg car possessing 3,000 kg(*)m/s of momentum?
Trava [24]

The magnitude of velocity for this car is equal to 1.5 m/s.

<u>Given the following data:</u>

  • Momentum of car = 3,000 kgm/s.
  • Mass of car = 2,000 kg.

To calculate the magnitude of velocity for this car:

<h3>What is momentum?</h3>

In Science, momentum simply means a multiplication of the mass of an object and its velocity.

Mathematically, momentum is giving by the formula;

Momentum = mass \times velocity

Making velocity the subject of formula, we have:

Velocity=\frac{Momentum}{Mass}

Substituting the given parameters into the formula, we have;

Velocity=\frac{3000}{2000}

Velocity = 1.5 m/s.

Read more on momentum here: brainly.com/question/15517471

5 0
3 years ago
PLEASE HELP!!!! WILL GIVE BRAINLIEST!!!!
Svetach [21]
There’s no image ???
4 0
3 years ago
Help this was due yesterday...
insens350 [35]

Answer:

Concept: Periodic Trends

  1. So we have Li and we have C
  2. Li has is lower in the periodic table and as you progress from s group to p group and eventually d group, the traction and ability to attract electrons increases.
  3. Hence option D
3 0
3 years ago
Read 2 more answers
A skier moving at 4.75 m/s encounters a long, rough, horizontal patch of snow having a coefficient of kinetic friction of 0.220
disa [49]
First we need to find the acceleration of the skier on the rough patch of snow.
We are only concerned with the horizontal direction, since the skier is moving in this direction, so we can neglect forces that do not act in this direction. So we have only one horizontal force acting on the skier: the frictional force, \mu m g. For Newton's second law, the resultant of the forces acting on the skier must be equal to ma (mass per acceleration), so we can write:
ma=-\mu m g
Where the negative sign is due to the fact the friction is directed against the motion of the skier.
Simplifying and solving, we find the value of the acceleration:
a=-(0.220)(9.81 m/s^2)=-2.16 m/s^2

Now we can use the following relationship to find the distance covered by the skier before stopping, S:
2aS=v_f^2-v_i^2
where v_f=0 is the final speed of the skier and v_i=4.75 m/s is the initial speed. Substituting numbers, we find:
S=- \frac{v_i^2}{2a}=- \frac{(4.75 m/s)^2}{2(-2.16 m/s^2)}=5.23 m
5 0
3 years ago
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