List six possible valve defects that should be included in the inspection of a used valve?

(a) Define a slip system. (b) Cite one slip system for the simple cubic crystal structure. (c) One slip system for the BCC cryst

Lemur [1.5K]

Answer:

I'm gonna say D.

4 0

3 years ago

In a 1D compression test with double drainage, the pore pressure readings are practically zero after 8 minutes for a clay sample

frosja888 [35]

Answer:

The duration of the consolidation process for the same clay is 32 min

Explanation:

for clay 1:

t1=0

H1=thickness=2 cm

for the clay 2:

t2=?

H2=2 cm

The time factor is equal to:

$T=(\frac{Cv}{d^{2} })t$

where Cv is the coefficient of consolidation

$(\frac{Cvt}{d^{2} })_{1}= (\frac{Cvt}{d^{2} })_{2}$

if Cv is constant, we have:

$(\frac{t1}{(\frac{H1}{2}) ^{2} })_{1}=(\frac{t2}{H2^{2} })_{2}\\\frac{0}{(\frac{2}{2})^{2}) }=\frac{t2}{2^{2} }$

Clearing t2:

t2=32 min

3 0

3 years ago

A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm (0.60 in. × 0.75 in.) is pulled in tension with 44,500

lukranit [14]

Answer:

The resulting strain is $1.39\times 10^{-3}$ .

Explanation:

A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm

Force, F = 44,500 N

Th elastic modulus of Cu to be 110 GPa

The resulting strain is given by the formula as follows :

$\epsilon=\dfrac{F}{AE}$

E is elastic modulus of Cu is are of cross section

$\epsilon=\dfrac{44500}{15.2\times 19.1\times 10^{-6}\times 110\times 10^9}\\\\\epsilon=1.39\times 10^{-3}$

So, the resulting strain is $1.39\times 10^{-3}$ .

6 0

4 years ago

The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta

tatyana61 [14]

Answer:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

$H_o =r x mv=rxL$

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

MO, where MO is the moment of the force F about point O. The equation expressing the rate of change of angular momentum is this one:

MO = H˙ O

Principle of Angular Impulse and Momentum

The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

$\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1$

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is $I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2$ ".