A change in ________ results in _______

A rod has a radius of 10 mm is subjected to an axial load of 15 N such that the axial strain in the rod is ????௫ = 2.75*10-6, de

EleoNora [17]

Answer:

Knowing we only have one load applied in just one direction we have to use the Hooke's law for one dimension

ex = бx/E

бx = Fx/A = Fx/π $r^{2}$

Using both equation and solving for the modulus of elasticity E

E = бx/ex = Fx / π $r^{2}$ ex

E = $\frac{15}{pi (10 * 10^{-3})^{2} * 2.75 * 10^{-6} } = 17.368 * 10^{9} Pa = 17.4 GPa$

Apply the Hooke's law for either y or z direction (circle will change in every direction) we can find the change in radius

ey = $\frac{1}{E}$ (бy - v (бx + бz)) = $-\frac{v}{E}$ бx

= $\frac{vFx}{Epir^{2} }$ = $\frac{0.23 * 15}{pi (10 * 10^{-3)^{2} } * 17.362 * 10^{9} } = -0.63 *10^{-6}$

Finally

ey = Δr / r

Δr = ey * r = 10 * $-0.63* 10^{-6} mm = -6.3 * 10^{-6} mm$

Δd = 2Δr = $-12.6 * 10^{-6} mm$

Explanation:

5 0

3 years ago

100POINTTTTTSSS PLEASE HELP

attashe74 [19]

Components connected in series are connected along a single path, so the same current flows through all of the components. If the light bulbs are connected in parallel, the currents through the light bulbs combine to form the current in the battery, while the voltage drop is across each bulb and they all glow.

8 0

4 years ago

A 65kg person throw a 0.045kg snowball forward with a ground speed of 30m/s. A second person, with a mass of 60kg, catches the s

Kobotan [32]

Well, st first we should find <span>initial momentum for the first person represented in the task which definitely must be :
</span> $(65+0.045)*2.5$
And then we find the final one : $65*x + 0.045*30$
Then equate them together : $x=2.48 m/s$
So we can get the velocity, which is $is 2.48 m/s$
In that way, according to the main rules of <span>conservation of momentum you can easily find the solution for the second person.
Regards!</span>

6 0

4 years ago

Pls help A car starts from rest and gains a velocity of 20m/s in 10 seconds calculate its acceleration and the distance covered

Soloha48 [4]

Answer:

$\boxed{\sf Acceleration \ (a) = 2 \ m/s^{2}}$