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maria [59]
2 years ago
10

What are limits that would be age appropriate for teens vs toddlers? give me some examples of limits that you may have for a tee

n vs toddller?
Physics
1 answer:
Pavlova-9 [17]2 years ago
6 0

The limits which would be age appropriate for toddlers include being flexible and realistic while that of a teen should be more rigid.

<h3>Who are Toddlers?</h3>

This refers to a child whose age ranges from 1 year to 4 years and limits set for them should be flexible due to it helping them feel more secure and decreases anxiety.

Example of limits that you may have for a teen is reducing their screen time while an example of limits for toddlers is correcting their junk eating habits. Teens should have a rigid limit so as to enable them do the right things at all times due to their strong emotions.

Read more about Limits here brainly.com/question/1419949

#SPJ1

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7 0
3 years ago
A mountain climber weighs 42.0 N. If he climbs a hill 100m high. Calculate the work done in joules​
quester [9]

Answer:

The answer is 4200 J.

Explanation:

The formula of work done is, W = F×D where F is the force of an object and D is the distance. Then you just substitute the values into the equation :

W = F×D

F = 42N

D = 100m

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= 4200 J

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nikitadnepr [17]

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3 years ago
According to the text, there is no energy shortage now, nor will there ever be. what reason (s) is given to support this stateme
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8 0
3 years ago
a mass of 0.75 kg is attached to a spring and placed on a horizontal surface. the spring has a spring constant of 180 N/m, and t
Artemon [7]

Answer:

6.57 m/s

Explanation:

First use Hook's Law to determine the F the compressed spring acts on the mass. Hook's Law F=kx; F=force, k=stiffnes of spring (or spring constant), x=displacement

F=kx; F=180(.3) = 54 N

Next from Newton's second law find the acceleration of the mass.

Newton's .2nd law F=ma; a=F/m ; a=54/.75 = 72m/s²

Now use the kinematic equation for velocity (or speed)

v₂²= v₀² + 2a(x₂-x₀); v₂=final velocity; v₀=initial velocity; a=acceleration; x₂=final displacement; x₀=initial displacment.

v₀=0, since the mass is at rest before we release it

a=72 m/s² (from above)

x₀=0 as the start position already compressed

x₂=0.3m (this puts the spring back to it's natural length)

v₂²= 0 + 2(72)(0.3) = 43.2 m²/s²

v₂=\sqrt{43.2)\\ = 6.57 m/s

5 0
3 years ago
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