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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 <u>actual</u> answer would be A: Regularly engage in weight-bearing activities

Explanation:

Evidence can be seen in this article about osteoporosis-

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A competitive go-cart driver is traveling at a speed of 32m/s. He sees a caution flag go up and slows down at a rate of -1.5 m/s

djyliett [7]

Answer:

His final velocity is 15.8 m/s.

Step-by-step explanation:

Given:

Initial velocity of the driver is, $u=32$ m/s

Acceleration of the driver is, $a=-1.5$ m/s²

Time taken to reach final velocity is, $t=10.8$ s.

The final velocity is given using the Newton's equations of motion as:

$v=u+at$ , where, $v$ is the final velocity.

Now, plug in the given values and solve for $v$ .

$v=32-1.5(10.8)\\v=32-16.2=15.8\textrm{ m/s}$

Therefore, his final velocity is 15.8 m/s.

5 0

3 years ago

a baseball is pitched with a speed of 35.0m/s. how long does it take the ball to travel 18.4 m from the pitchers mound to home p

Vlad [161]

Given the speed and the distance, to find time you can use the formula speed is equal to distance over time. From there you can manipulate the equation for time to equal the distance divided by speed. Time is equal to 18.4 meters divided by 35m/s which equals 0.526 seconds.

7 0

3 years ago

A small airplane has to reach a speed if 27.8 m/s to takeoff. It can accelerate at 2.00 m/s^2. What is the minimal length of run

pickupchik [31]

Using the constant acceleration formula v^2 = u^2 + 2as, we can figure out that it would take a distance of 193.21m to reach 27.8m/s

3 0

4 years ago

A 0.400-kg ice puck, moving east with a speed of 5.86 m/s , has a head-on collision with a 0.900-kg puck initially at rest.

andreev551 [17]

Answer:

a) The final speed of the 0.400-kg puck after the collision is 2.254 meters per second, b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards, c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

Explanation:

a) Since collision is perfectly elastic and there are no external forces exerted on pucks system, the phenomenon must be modelled after the Principles of Momentum and Energy Conservation. Changes in gravitational potential energy can be neglected. That is:

Momentum

$m_{1}\cdot v_{1,o} + m_{2}\cdot v_{2,o} = m_{1}\cdot v_{1,f} + m_{2}\cdot v_{2,f}$

Energy

$\frac{1}{2}\cdot (m_{1}\cdot v_{1,o}^{2}+ m_{2}\cdot v_{2,o}^{2})=\frac{1}{2}\cdot (m_{1}\cdot v_{1,f}^{2}+ m_{2}\cdot v_{2,f}^{2})$

$m_{1}\cdot v_{1,o}^{2} + m_{2}\cdot v_{2,o}^{2} = m_{1}\cdot v_{1,f}^{2} + m_{2}\cdot v_{2,f}^{2}$

Where:

$m_{1}$ , $m_{2}$ - Masses of the 0.400-kg and 0.900-kg pucks, measured in kilograms.

$v_{1,o}$ , $v_{2,o}$ - Initial speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

$v_{1}$ , $v_{2}$ - Final speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

If $m_{1} = 0.400\,kg$ , $m_{2} = 0.900\,kg$ , $v_{1,o} = +5.86\,\frac{m}{s}$ , $v_{2,o} = 0\,\frac{m}{s}$ , the system of equation is simplified as follows:

$2.344\,\frac{kg\cdot m}{s} = 0.4\cdot v_{1,f} + 0.9\cdot v_{2,f}$

$13.736\,J = 0.4\cdot v_{1,f}^{2}+0.9\cdot v_{2,f}^{2}$

Let is clear $v_{1,f}$ in first equation:

$0.4\cdot v_{1,f} = 2.344 - 0.9\cdot v_{2,f}$

$v_{1,f} = 5.86-2.25\cdot v_{2,f}$

Now, the same variable is substituted in second equation and resulting expression is simplified and solved afterwards:

$13.736 = 0.4\cdot (5.86-2.25\cdot v_{2,f})^{2}+0.9\cdot v_{2,f}^{2}$

$13.736 = 0.4\cdot (34.340-26.37\cdot v_{2,f}+5.063\cdot v_{2,f}^{2})+0.9\cdot v_{2,f}^{2}$

$13.736 = 13.736-10.548\cdot v_{2,f} +2.925\cdot v_{2,f}^{2}$

$2.925\cdot v_{2,f}^{2}-10.548\cdot v_{2,f} = 0$

$2.925\cdot v_{2,f}\cdot (v_{2,f}-3.606) = 0$

There are two solutions:

$v_{2,f} = 0\,\frac{m}{s}$ or $v_{2,f} = 3.606\,\frac{m}{s}$

The first root coincides with the conditions before collision and the second one represents a physically reasonable solution.

Now, the final speed of the 0.400-kg puck is: ( $v_{2,f} = 3.606\,\frac{m}{s}$ )

$v_{1,f} = 5.86-2.25\cdot (3.606)$

$v_{1,f} = -2.254\,\frac{m}{s}$

The final speed of the 0.400-kg puck after the collision is 2.254 meters per second.

b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards.

c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

3 0

3 years ago

The atoms in a solid move about freely

ivolga24 [154]

No, not exactly. They jiggle and tremble and vibrate a lot, but
they always basically stay in very nearly the same place.

It's like if you're allowed to go anywhere you want in your jail cell,
you wouldn't exactly call that "moving about freely".

6 0

3 years ago