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nekit [7.7K]
4 years ago
6

Which statement best describes the occurrence of maltreatment? ​

Physics
1 answer:
galina1969 [7]4 years ago
5 0

Answer:

b is the answer

b.There is a great deal of loyalty to the offenders. ​

Explanation:

There is a a little change in the rate of substantiated child maltreatment in the past several years.

In 2017 it is remarkably lower than 1990.

It has been decrease to 9 per 1000 children from 13 to 1000.

​ Since 2000 the rate of physical, psychological, sexual or emotional abuse have decreased .

The maltreatment rates are higher in younger children than older age children.

In maltreatment rate, children of ages 16 to 17 are lower than the children ages 0 to 3.

 The rate of maltreatment of smaller children is 3 times higher than bigger ones

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What was the most serious money problem faced by people in the early United States?
alexira [117]
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5 0
3 years ago
A student (m = 68 kg) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a
Gnesinka [82]

Answer:

5.7141 m

Explanation:

Here the potential and kinetic energy will balance each other

mgh=\frac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}

This is the initial velocity of the system and the final velocity is 0

t = Time taken = 0.04 seconds

F = Force = 18000 N

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

Equation of motion

v=u+at\\\Rightarrow a=\frac{v-u}{t}

From Newton's second law

F=ma\\\Rightarrow F=m\frac{v-u}{t}\\\Rightarrow 18000=68\frac{0-\sqrt{2gh}}{0.04}\\\Rightarrow \frac{18000}{68}\times 0.04=-\sqrt{2\times 9.81\times h}\\\Rightarrow 10.58823=-\sqrt{2\times 9.81\times h}

Squarring both sides

112.11061=2\times 9.81\times h\\\Rightarrow h=\frac{112.11061}{2\times 9.81}\\\Rightarrow h=5.7141\ m

The height from which the student fell is 5.7141 m

5 0
3 years ago
If the coefficient of kinetic friction between the puck and the ice is 0.05, how far does the puck slide before coming to rest?
Mariana [72]
According to newton's 3rd law of motion,

For every action, there is equal and opposite reaction. So if we move a body against a rough surface, there were be reaction against the force applied. 

So using conservation of energy, we know:

Work done to move a body = Work done against Friction
So, Force applied * distance moved = coefficient of Friction * Normal Reaction * distance moved

For a body moving against a normal surface, Normal Reaction (R) = mg

or, mass * acceleration * distance (s) = ∪ * R * distance(s)
or, mass * (v^2/2s) = ∪ * mass * gravity 

Now, s = stopping distance = v²/ 2∪g
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s = v2/2*0.05*g

We know, g = 10, so s = v²/(2*0.05*10) = v²

where v = initial velocity 
8 0
3 years ago
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Tema [17]

Answer:

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