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Kay [80]
3 years ago
6

Explain why 15yr old need more calcium than adults

Physics
2 answers:
Tom [10]3 years ago
8 0

Answer:

When adolescents get enough calcium during the teen years, they can start out their adult lives with the strong bones and significantly reduce their risk for fractures as an adult. Inadequate calcium intake during adolescence and young adulthood puts individuals at risk for developing osteoporosis later in life.

rodikova [14]3 years ago
8 0

Answer:

<h3> I hoped this helpful for you</h3>

Thank you ☺️☺️

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What is imperceptible human motion?
castortr0y [4]
Unable to be noticed or not felt because of feeling slight
3 0
3 years ago
What number is between 2000 and 2500
Kitty [74]
If its just a number 2,300 is an answer 
6 0
3 years ago
7. A toy car of mass 1.2 kg is driving vertical circles inside a hollow cylinder of radius 2.0m. It is moving at a constant spee
wlad13 [49]

Answer:

a)

N_{top}=9.8N\\N_{bottom}=33.4N

b) v_{min}=4.4m/s

Explanation:

The net force on the car must produce the centripetal acceleration necessary to make this circle, which is a_{cp}=\frac{v^2}{R}. At the top of the circle, the normal force and the weight point downwards (like the centripetal force should), while at the bottom the normal force points upwards (like the centripetal force should) and the weight downwards, so we have (taking the upwards direction as positive):

-m\frac{v^2}{R}=-N_{top}-mg\\m\frac{v^2}{R}=N_{bottom}-mg

Which means:

N_{top}=m\frac{v^2}{R}-mg=(1.2kg)\frac{(6m/s)^2}{2m}-(1.2kg)(9.8m/s^2)=9.8N\\N_{bottom}=m\frac{v^2}{R}+mg=(1.2kg)\frac{(6m/s)^2}{2m}+(1.2kg)(9.8m/s^2)=33.4N

The limit for falling off would be N_{top}=0, so the minimum speed would be:

0=m\frac{v_{min}^2}{R}-mg\\v_{min}=\sqrt{Rg}=\sqrt{(2m)(9.8m/s^2)}=4.4m/s

3 0
3 years ago
A spring with k = 15.3 N/cm is initially stretched 1.81 cm from its equilibrium length. a) How much more energy is needed to fur
leonid [27]

Answer:

2.31J

Explanation:

the energy for a spring system is given by:

E=\frac{1}{2} kx^2

where k is the spring constant: k=15.3N/cm=1530N/m and x is the distance stretched from the equilibrium position.

In the first case x=1.81cm=0.0181m

so the energy to stretch the spring 1.81cm is:

K_{1}=\frac{1}{2} (1530N/m)(0.0181m)^2=0.25J

and for the second case,  the energy to stretch the spring 5.79cm:

x=5.79cm=0.0579m

K_{1}=\frac{1}{2} (1530N/m)(0.0579m)^2=2.56J

so to answer a) we must find the difference between these energies:

2.56J-0.25J=2.31J

6 0
3 years ago
A big box of sausages (30 kg) is lifted from the ground to the top shelf of the freezer. If the box is lifted at a constant spee
Lubov Fominskaja [6]

Answer:

Work done to lift the box is 515.03 J

Explanation:

By work energy theorem we know that work done by all forces is equal to change in kinetic energy

So we have

W_g + W_{ex} = \Delta K

so we have

-mgh + W_{ex} = 0

so we have

W_{ex} = mgh

W_{ex} = 30(9.8)1.75

W_{ex} = 515.03 J

5 0
3 years ago
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