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zvonat [6]
3 years ago
13

what is the recommended daily dosage of flouride for a child over three years of age living in an area where drinking water is f

ree fluoride
Physics
1 answer:
Finger [1]3 years ago
7 0
<span>What is the recommended daily dosage of flouride for a child over three years of age living in an area where drinking water is free fluoride</span>child:

3-6: 0.5mg 
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Worth 15 points<br><br> Please answer which one matches for each conversion
FrozenT [24]

Answer:

distance - meters

speed - meters/seconds

time - seconds

velocity - meters/seconds

acceleration - meters/seconds²

5 0
2 years ago
A 5 kilograms bowling ball is dropped out a window. It hits the ground, and bounces upward. The velocity change of the ball is n
ioda

Answer:

13.5

Explanation:

Mass: 5kg

Initial Velocity: -15

Final Velocity: 12

Force: 10

We can use the equation: Vf = Vi + at

We need to find acceleration, and we can use the equation, F=ma,

We have mass and the force so it would look like this, 10=5a, and 5 times 2 would equal 10, so acceleration would be 2.

Now we have all the variables to find time.

Back to Vf = Vi + at, plug the numbers in, 12 = -15 + 2(t)

Plugging them in into desmos gives 13.5 for time.

4 0
2 years ago
How to calculate F2?<br> m=16.4kg<br> f1= 2.7n<br> angle=34.4
V125BC [204]
Option 2 is your answer :)
4 0
3 years ago
Read 2 more answers
Time shifting occurs when _______.
Svet_ta [14]
The answer is C, individuals copy works to view at a later time.
4 0
3 years ago
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A horse of mass 242 kg pulls a cart of mass 224 kg. The acceleration of gravity is 9.8 m/s 2 . What is the largest acceleration
Rufina [12.5K]

To solve this problem it is necessary to apply the concepts related to Newton's second Law and the force of friction. According to Newton, the Force is defined as

F = ma

Where,

m= Mass

a = Acceleration

At the same time the frictional force can be defined as,

F_f = \mu N

Where,

\mu = Frictional coefficient

N = Normal force (mass*gravity)

Our values are given as,

m_h = 242 kg\\m_c = 224 kg\\\mu = 0.894\\

By condition of Balance the friction force must be equal to the total net force, that is to say

F_{net} = F_f

m_{total}a = \mu m_hg

(m_h+m_c)a = \mu*m_h*g

Re-arrange to find acceleration,

a= \frac{\mu*m_h*g}{(m_h+m_c)}

a = \frac{0.894*242*9.8}{(242+224)}

a = 4.54 m/s^2

Therefore the acceleration the horse can give is 4.54m/s^2

3 0
3 years ago
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