Fastest
(Known as the fast lane)
Answer:
the elongation of the metal alloy is 21.998 mm
Explanation:
Given the data in the question;
K = σT/ (εT)ⁿ
given that metal alloy true stress σT = 345 Mpa, plastic true strain εT = 0.02,
strain-hardening exponent n = 0.22
we substitute
K = 345 / ![0.02^{0.22](https://tex.z-dn.net/?f=0.02%5E%7B0.22)
K = 815.8165 Mpa
next, we determine the true strain
(εT) = (σT/ K)^1/n
given that σT = 412 MPa
we substitute
(εT) = (412 / 815.8165 )^(1/0.22)
(εT) = 0.04481 mm
Now, we calculate the instantaneous length
= ![l_0e^{ET](https://tex.z-dn.net/?f=l_0e%5E%7BET)
given that
= 480 mm
we substitute
=
× ![e^{0.04481](https://tex.z-dn.net/?f=e%5E%7B0.04481)
= 501.998 mm
Now we find the elongation;
Elongation = ![l_i - l_0](https://tex.z-dn.net/?f=l_i%20-%20l_0)
we substitute
Elongation = 501.998 mm - 480 mm
Elongation = 21.998 mm
Therefore, the elongation of the metal alloy is 21.998 mm
Answer:All of the above
Explanation:
9.62 psi means 497.49 mm of Hg pressure
for (a)19.58 inches is equals to 497.49 mm of Hg
(b)atmospheric pressure is 14.69 psi
vaccum gauge is 9.62psi
absolute pressure is=14.69-9.62=5.07
(c)vaccum means air is sucked and there is negative pressure so it tells about below atmospheric pressure.
thus all are correct
Answer:
Carpenter's square
Explanation:
The most common hand tool used to measure or set angles with its application extending to setting angles of roofs and rafters. Another name of a Carpenter's square is a framing square.
Other hand tools that are used to measure angles are;
- The combination square that allows a user to set both 90° and 45° angles
- A Bevel that allows users to set any angle they like.
- A Protractor that resembles a bevel but its marks are marked in an arc.
- An electromagnetic angle finder which gives a reading according to the measure of the arms adjusted by the user.
Given:
diameter of sphere, d = 6 inches
radius of sphere, r =
= 3 inches
density,
= 493 lbm/ ![ft^{3}](https://tex.z-dn.net/?f=ft%5E%7B3%7D)
S.G = 1.0027
g = 9.8 m/
= 386.22 inch/ ![s^{2}](https://tex.z-dn.net/?f=s%5E%7B2%7D)
Solution:
Using the formula for terminal velocity,
=
(1)
![(Since, m = V\times \rho)](https://tex.z-dn.net/?f=%28Since%2C%20m%20%3D%20V%5Ctimes%20%5Crho%29)
where,
V = volume of sphere
= drag coefficient
Now,
Surface area of sphere, A = ![4\pi r^{2}](https://tex.z-dn.net/?f=4%5Cpi%20r%5E%7B2%7D)
Volume of sphere, V = ![\frac{4}{3} \pi r^{3}](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20r%5E%7B3%7D)
Using the above formulae in eqn (1):
= ![\sqrt{\frac{2\times \frac{4}{3} \pir^{3}\rho g}{4\pi r^{2} \rho C_{d}}}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B2%5Ctimes%20%5Cfrac%7B4%7D%7B3%7D%20%5Cpir%5E%7B3%7D%5Crho%20%20g%7D%7B4%5Cpi%20r%5E%7B2%7D%20%5Crho%20C_%7Bd%7D%7D%7D)
=
= ![\sqrt{\frac{2\times 386.22\times 3}{3C_{d}}}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B2%5Ctimes%20386.22%5Ctimes%203%7D%7B3C_%7Bd%7D%7D%7D)
Therefore, terminal velcity is given by:
=
inch/sec