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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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Bromium has two naturally occurring isotopes: 79br, with an atomic weight of 78.918 amu, and 81br, with an atomic weight of 80.9
saul85 [17]

The two different isotopes have weights :

w1 = 78.918 amu

w2 = 80.916 amu

average weight w3 = 79.903 amu

The mixing of two components can be modeled as

let the fraction of w1 be 'x'

hence w1. x + w2.(1-x)  = w3

now this is a linear equation in 'x'. Substituting the values we get

x = 0.507

hence the percentage of Br79 = 50.7% and the percentage of BR81 = 49.3%

8 0
3 years ago
At the same instant that a 0.50-kg ball is dropped from 25m above Earth,? At the same instant that a 0.50-kg ball is dropped fro
fiasKO [112]
I hope it is clearly visible.. Velocity of the center of mass of 2-ball system is - 11.54m/s. Minus indicates, velocity direction is in downward direction.

6 0
3 years ago
When the Moon orbits Earth, what is the centripetal force?
Troyanec [42]

Answer:

D

Explanation:

pls mark brainliest

7 0
3 years ago
A man does 4,475 J of work in the process of pushing his 2.50 103 kg truck from rest to a speed of v, over a distance of 26.0 m.
Tcecarenko [31]

Answer:

a) 1.89 m/s  b) 172.1 N

Explanation:

a)

  • Applying the work-energy theorem, if we can neglect the friction between truck and road, the total change in kinetic energy must be equal to the work done by the external forces.
  • This work, is just 4,475 J.
  • So we can write the following equation:

        \Delta K = \frac{1}{2} * m*v^{2} = 4,475 J

  • where m= mass of the truck = 2.5*10³ kg.
  • So, we can find the speed v, as follows:

        v =\sqrt{\frac{2*W}{m}} =\sqrt{\frac{2*4,475J}{2.5e3kg} }  = 1.89 m/s

b)

  • The work done by the man, is just the horizontal force applied, times the displacement produced by the force horizontally:

        W = F*d

  • We can solve for F, as follows:

        F = \frac{W}{d} = \frac{4,475 J}{26.0m} =  172.1 N

4 0
3 years ago
A cylinder of mass 14.0 kg rolls without slipping on a horizontal surface. At a certain instant its center of mass has a speed o
aev [14]

Answer:

a) 567J

b) 283.5J

c)850.5J

Explanation:

The expression for the translational kinetic energy is,

E_r = \frac{1}{2} mv^2

Substitute,

14kg for m

9m/s for v

E_r = \frac{1}{2} (14) (9)^2\\= 567J

The translational kinetic energy of the center of mass is 567J

(B)

The expression for the rotational kinetic energy is,

E_R = \frac{1}{2} Iw^2

The expression for the moment of inertia of the cylinder is,

I = \frac{1}{2} mr^2

The expression for angular velocity is,

w = \frac{v}{r}

substitute

1/2mr² for I

and vr for w

in equation for rotational kinetic energy as follows:

E_R = (\frac{1}{2}) (\frac{1}{2} mr^2)(\frac{v}{r} )^2

= \frac{mv^2}{4}

E_R = \frac{14 \times 9^2 }{4} \\\\= 283.5J

The rotational kinetic energy of the center of mass is 283.5J

(c)

The expression for the total energy is,

E = E_r + E_R\\\\

substitute 567J for E(r) and 283.5J for E(R)

E = 567J + 283.5\\= 850.5J

The total energy of the cylinder is 850.5J

6 0
3 years ago
Read 2 more answers
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