Ask question

Ask question

All categories

2 years ago

7

A 85-kg person stands on one leg and 90% of the weight is supported by the upper leg connecting the knee and hip joint – the fem

ur. Assuming the femur is 0.65 m long and has a radius of 2.00 cm, by how much will the bone be compressed (in meters)?

1 answer:

Maru [420]2 years ago

4 0

Answer:

The bone is compressed to 4.213*10⁻⁶ m.

Explanation:

Given that ,

Mass of the person M = 85 kg,

Length of femur L =0.65 m

Radius of femur r = 0.02 m

To find,

Compression of the bone l = ?

The Young's Modulus is given by Y= $\frac{FL}{lA}$

Young's modulus of human femur bone is Y = 9.4 *10⁹ Pa

Rearranging formula ,

l = $\frac{FL}{YA}$

Weight supported by femur bone is 85 *0.9 =76.5 Kg

Area = π r² =3.14* (0.02)² = 0.001256 m

The length of compressed bone is

l = $\frac{76.5*0.65}{9.4GPa*0.001256}$ = $\frac{49.725}{0.0118GPa}$ =4213.89 *10⁻⁹

l = 4.213*10⁻⁶ m

The bone is compressed to 4.213*10⁻⁶ m.

You might be interested in

Two trucks with equal mass are attracted to each other with a gravitational force of 5.3 x 10 -4 N. The trucks are separated by

HACTEHA [7]

Answer:

7327 kg or 7.3 tons

Explanation:

We have Newton formula for attraction force between 2 objects with mass and a distance between them:

$F_G = G\frac{M_1M_2}{R^2}$

where $G =6.67408*10^{-11} m^3/kgs^2$ is the gravitational constant on Earth. $M = M_1 = M_2$ is the masses of the 2 objects. and R = 2.6m is the distance between them.

$F_G = 5.3*10^{-4}N$ is the attraction force.

$5.3*10^{-4} = 6.67408*10^{-11}\frac{M^2}{2.6^2}$

$7941265 = \frac{M^2}{2.6^2}$

$M^2 = 53682948.76$

$M = \sqrt{53682948.76} \approx 7327 kg$ or 7.3 tons

3 0

2 years ago

Which best describes what forms in nuclear fission?A. two smaller, more stable nucleiB. two larger, less stable nucleiC. one sma

Step2247 [10]

Answer:B. two larger, less stable nuclei

Explanation: They collied and don't combine

4 0

2 years ago

Read 2 more answers

A person on a cruise ship is doing laps on the promenade deck. on one portion of the track the person is moving north with a spe

Dmitry [639]

The resultant motion is given by pithagoras, since the two components (north and east) are perpendicular to each other.
They are asking you about the direction so you have to use trigonometry, finding that the direction is Ф=arctan(3.8/12)=17.57° north of east.

4 0

2 years ago

Read 2 more answers

What is the rate of heat transfer by radiation, with an unclothed person standing in a dark room whose ambient temperature is 22

SIZIF [17.4K]

Answer:

$5.45\times 10^{-4}$ W

Explanation:

$T_{r}$ = Temperature of the room = 22.0 °C = 22 + 273 = 295 K

$T_{s}$ = Temperature of the skin = 33.0 °C = 33 + 273 = 306 K

$A$ = Surface area = 1.50 m²

$\epsilon$ = emissivity = 0.97

$\sigma$ = Stefan's constant = 5.67 x 10⁻⁸ Wm⁻² K⁻⁴

Rate of heat transfer is given as

$R = \epsilon \sigma A (T_{s}^{2} - T_{r}^{2})$

$R = (0.97)(5.67\times 10^{-8}) (1.50) ((306)^{2} - (295)^{2})$

$R = 5.45\times 10^{-4}$ W

3 0

2 years ago

How much energy is needed to change the speed of a 1600 kg sport utility vehicle from 15.0 m/s to 40.0 m/s?

podryga [215]

Answer:

Energy needed = 1100 kJ

Explanation:

Energy needed = Change in kinetic energy

Initial velocity = 15 m/s

Mass, m = 1600 kg

$\texttt{Initial kinetic energy = }\frac{1}{2}mv^2=0.5\times 1600\times 15^2=180000J$

Final velocity = 40 m/s

$\texttt{Final kinetic energy = }\frac{1}{2}mv^2=0.5\times 1600\times 40^2=1280000J$

Energy needed = Change in kinetic energy = 1280000-180000 = 1100000J

Energy needed = 1100 kJ

4 0

2 years ago