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2 years ago

The size of an engine is called the engine

Engineering

2 answers:

hjlf2 years ago

6 0

What’s there too answer if you answered it

harina [27]2 years ago

3 0

So what am supposed to answer

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What do you mean by searching?

nikitadnepr [17]

Answer:

thoroughly scrutinizing, especially in a disconcerting way.

Explanation:

8 0

3 years ago

You are given a body with no body forces and told that the stress state is given as: ⎡ ⎣ 3αx 5βx2 + αy γz3 5βx2 + αy βx2 0 γz3 0

Lady bird [3.3K]

Answer:

This doesn't represent an equilibrium state of stress

Explanation:

∝ = 1 , β = 1 , y = 1

x = 0 , y = 0 , z = 0 ( body forces given as 0 )

Attached is the detailed solution is and also the conditions for equilibrium

for a stress state to be equilibrium all three conditions has to meet the equilibrum condition as explained in the attached solution

5 0

3 years ago

A polyethylene rod exactly 10 inches long with a cross-sectional area of 0.04 in2 is used to suspend a weight of 358 lbs-f (poun

Nadya [2.5K]

Answer:

Final length of the rod = 13.90 in

Explanation:

Cross Sectional Area of the polythene rod, A = 0.04 in²

Original length of the polythene rod, l = 10 inches

Tensile modulus for the polymer, E = 25,000 psi

Viscosity, $\eta = 1*10^{9} psi -sec$

Weight = 358 lbs - f

time, t = 1 hr = 3600 sec

Stress is given by:

$\sigma = \frac{Force}{Area} \\\sigma = \frac{358}{0.04} \\\sigma = 8950 psi$

Based on Maxwell's equation, the strain is given by:

$strain = \sigma ( \frac{1}{E} + \frac{t}{\eta} )\\Strain = 8950 ( \frac{1}{25000} + \frac{3600}{10^{9} } )\\Strain = 0.39022$

Strain = Extension/(original Length)

0.39022 = Extension/10

Extension = 0.39022 * 10

Extension = 3.9022 in

Extension = Final length - Original length

3.9022 = Final length - 10

Final length = 10 + 3.9022

Final length = 13.9022 in

Final length = 13.90 in

7 0

3 years ago

Determine the voltages at all nodes and the currents through all branches. Assume that the transistor B is 100,

iren [92.7K]

Answer:

The voltages of all nodes are, IE = 4.65 mA, IB =46.039μA, IC=4.6039 mA, VB = 10v, VE =10.7, Vc =4.6039 v

Explanation:

Solution

Given that:

V+ = 20v

Re = 2kΩ

Rc = 1kΩ

Now we will amke use of the method KVL in the loop.

= - Ve + IE . Re + VEB + VB = 0

Thus

IE = V+ -VEB -VB/Re

Which gives us the following:

IE = 20-0.7 - 10/2k

= 9.3/2k

so, IE = 4.65 mA

IB = IE/β +1 = 4.65 m /101

Thus,

IB = 0.046039 mA

IB = 46.039μA

IC =βIB

Now,

IC = 100 * 0.046039

IC is 4.6039 mA

Now,

VB = 10v

VE = VB + VEB

= 10 +0.7 = 10.7 v

So,

Vc =Ic . Rc = 4.6039 * 1k

=4.6039 v

Finally, this is the table summary from calculations carried out.

Summary Table

Parameters IE IC IB VE VB Vc

Unit mA mA μA V V V

Value 4.65 4.6039 46.039 10.7 10 4.6039

4 0

2 years ago

A controlled process is described by the closed-loop transfer function G(s).

MissTica

Answer:

The answer is "Option B".

Explanation:

Given equation:

$G(s) =\frac{K(s + 1)}{2s^2 + (K-1)s + (K-1)}\\\\$

$\to 2s^2 + (K-1)s + (K-1)=0$

Calculating by the Routh's Hurwitz table:

$\to s^2 \ \ \ \ \ 2 \ \ \ \ \ \ K-1 \\\\\to s^2 \ \ \ \ \ K-1 \ \ \ \ \ \ \\\\\to s^0 \ \ ( \frac{(K-1)(K-1)(-2) (0)}{K-1} \\\\ \ \ \ \ = (K-1) )$

Form the above table:

$\to K-1 > 0 \\\\ \to K > 1$

In the above, the value of k is greater than 1.

3 0

3 years ago