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il63 [147K]
3 years ago
6

Which of the following is not an appropriate category of childrens book to include in the early childhood classroom?

Physics
2 answers:
erma4kov [3.2K]3 years ago
5 0
The answer is A because at a young age children will not understand and they need to learn at the appropriate level
matrenka [14]3 years ago
5 0

Answer:

chapter book

Explanation:

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P = 8667 W

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A 50kg skater at rest on a frictionless rink throws a 2kg ball, giving the ball a velocity of 10m/s. Which statement describes t
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0.4m/s in the opposite direction of the ball.

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3 years ago
URGENT <br> Find the total<br> equivalent<br> resistance for the<br> circuit.
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20 + 10 = 30 \\ 30and50 in \: parallel \\  =  \frac{50 \times 30}{50 + 30}  \\  =  \frac{1500}{80}  \\ 18.75 \\ 30nd40parallel \\  \frac{30 \times 40}{70}  \\  \frac{1200}{70}  = 17.143 \\ total = 18.75 + 17.143 = 35.893ohm \\ thank \: u

3 0
3 years ago
Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate
maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

       The length of the track is l =  1.0 \ km  =  1000 \  m

       The speed of  A is  v__{A}} =  20 \ m/s

       The uniform acceleration of  B is  a__{B}} =  0.333 \ m/s^2

  Generally the time taken by go-cart  A is mathematically represented as

              t__{A}} = \frac{l}{v__{A}}}

=>          t__{A}} = \frac{1000}{20}

=>           t__{A}} =  50 \  s

  Generally from kinematic equation we can evaluate the time taken by go-cart B as

             l =  ut__{B}} + \frac{1}{2}  a__{B}} * t__{B}}^2

given that go-cart B starts from rest  u =  0 m/s

So

            1000 =  0 *t__{B}} + \frac{1}{2}  * 0.333  * t__{B}}^2

=>         1000 =  0 *t__{B}} + \frac{1}{2}  0.333  * t__{B}}^2            

=>         t__{B}} =  77.5 \  seconds  

 

Comparing  t__{A}} \  and  \ t__{B}}  we see that t__{A}} is smaller so go-cart A is  faster

   

       

3 0
3 years ago
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