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olchik [2.2K]
1 year ago
9

Preferential stress will cause crystals to rotate into a preferred orientation unless the crystals are ________

Engineering
1 answer:
velikii [3]1 year ago
6 0

Answer:

Preferential stress will cause crystals to rotate into a preferred orientation unless the crystals are Equant

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Please write the following code in Python 3. Also please show all output(s) and share your code.
maksim [4K]

Answer:

sum2 = 0

counter = 0

lst = [65, 78, 21, 33]

while counter < len(lst):

   sum2 = sum2 + lst[counter]

   counter += 1

Explanation:

The counter variable is initialized to control the while loop and access the numbers in <em>lst</em>

While there are numbers in the <em>lst</em>,  loop through <em>lst</em>

Add the numbers in <em>lst</em> to the sum2

Increment <em>counter</em> by 1 after each iteration

6 0
3 years ago
Calculate the radius of gyration for a bar of rectangular cross section with thickness t and width w for bending in the directio
SVEN [57.7K]

Answer:

a)R= sqrt( wt³/12wt)

b)R=sqrt(tw³/12wt)

c)R= sqrt ( wt³/12xcos45xwt)

Explanation:

Thickness = t

Width = w

Length od diagonal =sqrt (t² +w²)

Area of raectangle = A= tW

Radius of gyration= r= sqrt( I/A)

a)

Moment of inertia in the direction of thickness I = w t³/12

R= sqrt( wt³/12wt)

b)

Moment of inertia in the direction of width I = t w³/12

R=sqrt(tw³/12wt)

c)

Moment of inertia in the direction of diagonal I= (w t³/12)cos 45=( wt³/12)x 1/sqrt (2)

R= sqrt ( wt³/12xcos45xwt)

4 0
3 years ago
The electric motor exerts a torque of 800 N·m on the steel shaft ABCD when it is rotating at a constant speed. Design specificat
kodGreya [7K]

Answer:

d= 4.079m ≈ 4.1m

Explanation:

calculate the shaft diameter from the torque,    \frac{τ}{r} = \frac{T}{J} = \frac{C . ∅}{l}

Where, τ = Torsional stress induced at the outer surface of the shaft (Maximum Shear stress).

r = Radius of the shaft.

T = Twisting Moment or Torque.

J = Polar moment of inertia.

C = Modulus of rigidity for the shaft material.

l = Length of the shaft.

θ = Angle of twist in radians on a length.  

Maximum Torque, ζ= τ ×  \frac{ π}{16} × d³

τ= 60 MPa

ζ= 800 N·m

800 = 60 ×  \frac{ π}{16} × d³

800= 11.78 ×  d³

d³= 800 ÷ 11.78

d³= 67.9

d= \sqrt[3]{} 67.9

d= 4.079m ≈ 4.1m

3 0
2 years ago
Read 2 more answers
Consider an aircraft powered by a turbojet engine that has a pressure ratio of 9. The aircraft is stationary on the ground, held
77julia77 [94]

Answer:

The break force that must be applied to hold the plane stationary is 12597.4 N

Explanation:

p₁ = p₂, T₁ = T₂

\dfrac{T_{2}}{T_{1}} = \left (\dfrac{P_{2}}{P_{1}}  \right )^{\frac{K-1}{k} }

{T_{2}}{} = T_{1} \times \left (\dfrac{P_{2}}{P_{1}}  \right )^{\frac{K-1}{k} } = 280.15 \times \left (9  \right )^{\frac{1.333-1}{1.333} } = 485.03\ K

The heat supplied = \dot {m}_f × Heating value of jet fuel

The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s

The heat supplied = \dot m · c_p(T_3 - T_2)

\dot m = 20 kg/s

The heat supplied = 20*c_p(T_3 - T_2) = 21,350 kJ/s

c_p = 1.15 kJ/kg

T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K

p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa

p₃ = p₂ = 855 kPa

T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K

T₄ = 1413.3 - 204.88 = 1208.42 K

\dfrac{T_5}{T_4}  = \dfrac{2}{1.333 + 1}

T₅ = 1208.42*(2/2.333) = 1035.94 K

C_j = \sqrt{\gamma \times R \times T_5} = √(1.333*287.3*1035.94) = 629.87 m/s

The total thrust = \dot m × C_j = 20*629.87 = 12597.4 N

Therefore;

The break force that must be applied to hold the plane stationary = 12597.4 N.

5 0
2 years ago
One good way to improve your gas Milage is to ___.
VashaNatasha [74]

Answer: B

Explanation:

One good way to improve your gas mileage is to accelerate smoothly and directly to a safe speed.

Hope this helps!

5 0
2 years ago
Read 2 more answers
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