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3 years ago

Bone has a Young's modulus of about

Physics

1 answer:

Oksanka [162]3 years ago

8 0

Answer:

505.62 mm

Explanation:

Young's modulus of a material is given by,

$Y=\frac{Stress}{Strain}$

Stress is the force experienced by the material per unit area (which is why stress has the same units as pressure).

Strain is the ratio between the change in length/area/volume to the original length/area/volume of the material (and hence a dimensionless quantity). Basically, it gives us a measure of the change in dimensions due to the stress experienced.

So, for the given problem,

$Y=\frac{Stress}{\Delta L/L}$

or, $1.8\times100=\frac{1.59\times108}{\Delta L/0.53}$ (since we need to find the compression of the bone at the critical stress point)

or, $\Delta L=\frac{1.59\times108\times0.53}{1.8\times100} m=0.50562m=505.62mm$

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The fluid pressure 10 ft underwater is ____ the fluid pressure 5 ft underwater

quester [9]

5ft is two times the 10ft

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3 years ago

A boat is subject to hydrodynamic drag forces of F = 40v 2 [N] where v is the magnitude of velocity in [m/s]. The boat’s mass is

stepladder [879]

Answer:

It has moved a distance, S = 25.9 m

Explanation:

F = 40 v²........(1)

$F = mv\frac{dv}{dS}$ .........(2)

Equating (1) and (2)

$-mv\frac{dv}{dS} = 40 v^2\\\\v\frac{dv}{dS} = \frac{-40 v^2}{m} \\\\m = 450 kg\\v\frac{dv}{dS} = \frac{-40 v^2}{450}\\\\ \frac{dv}{v} = \frac{40}{450} dS\\$

Integrate both sides:

$v_1 = 1, v_2 = 10$

$\int\limits^{10}_1 {\frac{1}{v} } \, dv = \frac{-40}{450} \int\limits^S_0 \, dS \\\\ln\frac{1}{10} = \frac{-40}{450} (S-0)\\\\S = \frac{-40}{450} ln(0.1)\\\\S = 25.9 m$ .

6 0

3 years ago

What is the force of gravitational attraction between an object with a mass of 100 kg and another object that has a mass of 300

Leokris [45]

Answer:

5 x 10⁻⁷N

Explanation:

Given parameters:

Mass of object 1 = 100kg

Mass of object 2 = 300kg

Distance = 2m

Unknown:

Force of gravitational attraction between the objects = ?

Solution:

Newton's law of universal gravitation states that the gravitational force between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

From Newton's law of universal gravitation we derive an expression:

Fg = $\frac{Gm_{1} m_{2} }{r^{2} }$

G is the universal gravitation constant = 6.67 x 10⁻¹¹

m is the mass

r is the distance between the bodies

Now insert the parameters and solve;

Fg = 6.67 x 10⁻¹¹ x $\frac{100 x 300}{2^{2} }$ = 5 x 10⁻⁷N

7 0

3 years ago

A proton is launched from an infinite plane of charge with surface charge density -1.10×10-6 C/m2. If the proton has an initial

Katyanochek1 [597]

The distance covered by the proton is 48.4 m

Explanation:

The electric field produced by an electrically charged infinite plane is given by

$E=\frac{\sigma}{2\epsilon_0}$

where in this case,

$\sigma = -1.10\cdot 10^{-6} C/m^2$ is the surface charge density

$\epsilon_0 = 8.85\cdot 10^{-12} F/m$ is the vacuum permittivity

Substituting,

$E=\frac{-1.10\cdot 10^{-6}}{2(8.85\cdot 10^{-12})}=-5.65\cdot 10^4N/C$

And the direction is towards the plane (because the charge is negative).

The electric force on the proton due to this field is

$F=qE$

where

$q=1.6\cdot 10^{-19}C$ is the proton charge

Substituting,

$F=(1.6\cdot 10^{-19})(-5.65\cdot 10^4)=-9.0\cdot 10^{-15} N$

where the direction is toward the plane.

Now we can calculate the proton's acceleration using Newton's second law:

$a=\frac{F}{m}$

where

$m=1.67\cdot 10^{-27}kg$ is the proton mass

Substituting,

$a=\frac{-9.0\cdot 10^{-15}}{1.67\cdot 10^{-27}}=-5.4\cdot 10^{-12} m/s^2$

Now we can finally apply the following suvat equation for accelerated motion to find the distance travelled by the proton:

$v^2-u^2=2as$

where

v = 0 is the final velocity

$u=2.40\cdot 10^7 m/s$ is the initial velocity

a is the acceleration

s is the distance covered

And solving for s,

$s=\frac{v^2-u^2}{2a}=\frac{0-(2.4\cdot 10^7)^2}{2(-5.4\cdot 10^{12})}=53.3 m$

Therefore, the closest answer is 48.4 m.

Learn more about accelerated motion:

brainly.com/question/9527152

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brainly.com/question/2562700

#LearnwithBrainly

5 0

3 years ago

What is the acceleration of a proton moving with a speed of 7.7 m/s at right angles to a magnetic field of 1.9 T

iris [78.8K]

Answer:

The acceleration of the proton is 1.403 x 10⁹ m/s²

Explanation:

Given;

speed of proton, v = 7.7 m/s

magnitude of magnetic field, B = 1.9 T

Magnetic force of moving proton is given by;

F = qvBsinθ

Centripetal force on the moving proton is given by;

$F = m(\frac{v^2}{r})\\\\F = m(a_c) \\\\a_c \ is \ the \ centripetal \ acceleration$

$qvBsin\theta = ma_c\\\\ac = \frac{qvBsin(90)}{m}$

where;

q is charge of the proton = 1.602 x 10⁻¹⁹ C

m is mass of proton = 1.67 x 10⁻²⁷ kg

$ac = \frac{(1.602*10^{-19})(7.7)(1.9)sin(90)}{1.67*10^{-27}}\\\\a_c = 1.403*10^{9} \ m/s^2$

Therefore, the acceleration of the proton is 1.403 x 10⁹ m/s²

3 0

3 years ago