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3 years ago

What should a person do to reduce the risk of developing osteoporosis if it runs in the family?

Physics

1 answer:

Cerrena [4.2K]3 years ago

8 0

Answer:

The actual answer would be A: Regularly engage in weight-bearing activities

Explanation:

Evidence can be seen in this article about osteoporosis-

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How long will your trip take if you travel 4000 m at an average speed of 8m/s

SpyIntel [72]

Answer:

500 sec

8 min 20 sec

Explanation:

Hi there !

8 m ................ 1 s

4000 m ........ x s

x = 4000m×1s/8m = 500 sec = 8 min 20 sec

Good luck !

8 0

3 years ago

Read 2 more answers

Flasher units are being discussed. Technician A says that only a DOT-approved flasher unit should be used for turn signals. Tech

zmey [24]

Answer: C

Both Technicians A and B

Explanation:

Only a DOT-approved flasher unit should be used for turn signals. And a parallel (variable-load) flasher will function for turn signal usage, although it will not warn the driver if a bulb burns out.

3 0

3 years ago

Meme on newtons law of motion (should me made ur self not from any searh engine)

k0ka [10]

He will be a pilot and he will fly the plane over bridges fewwww

8 0

1 year ago

Please help!!!

Zolol [24]

Answer:

Explanation:

you basically divide 1200 into 25

8 0

3 years ago

A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so tha

Dvinal [7]

Answer:

0.087 m

Explanation:

Length of the rod, L = 1.5 m

Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.

time period, T = 3 s

the formula for the time period of the pendulum is given by

$T = 2\pi \sqrt{\frac{I}{mgd}}$ .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

Moment of inertia of the rod about the centre of mass, Ic = mL²/12

By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

I = Ic + md²

$I = \frac{mL^{2}}{12}+ md^{2}$

Substituting the values in equation (1)

$3 = 2 \pi \sqrt{\frac{\frac{mL^{2}}{12}+ md^{2}}{mgd}}$

$9=4\pi^{2}\times \left ( \frac{\frac{L^{2}}{12}+d^{2}}{gd} \right )$

12d² -26.84 d + 2.25 = 0

$d=\frac{26.84\pm \sqrt{26.84^{2}-4\times 12\times 2.25}}{24}$

$d=\frac{26.84\pm 24.75}{24}$

d = 2.15 m , 0.087 m

d cannot be more than L/2, so the value of d is 0.087 m.

Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.

3 0

3 years ago