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nadya68 [22]
3 years ago
12

Social engineering

Engineering
1 answer:
Nat2105 [25]3 years ago
7 0

Answer:

withdraw

blow

view

review

draw

throw

show

interview

allow

grow

know

borrow

narrow

renew

flow

follow

bow

swallow

slow

Explanation:

withdraw

blow

view

review

draw

throw

show

interview

allow

grow

know

borrow

narrow

renew

flow

follow

bow

swallow

slow

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4. Which of the following drive roll designs is used
Yuki888 [10]
Knurled drive rolls are used with gas and self shielded flux cored and metal cored wires, which are softer due to the flux inside and the tubular design.
5 0
4 years ago
Air enters the compressor of an ideal gas refrigeration cycle at 7∘C and 35 kPa and the turbine at 37∘C and 160 kPa. The mass fl
amid [387]

It appears that your answer contains either a link or inappropriate words. Please correct and submit again! error

Had to screenshot the solution check attached

5 0
4 years ago
Consider the base plate of an 800-W household iron with a thickness of L 5 0.6 cm, base area of A 5 160 cm2, and thermal conduct
N76 [4]

Answer:

a. \frac{-kdT(0)}{dx} =q_{0}=5000W/m^2

b.833.3(0.006-x)+112

c. 117 deg C

Explanation:

Consider the base plate of an 800-W household iron with a thickness of L 5 0.6 cm, base area of A 5 160 cm2, and thermal conductivity of k 5 60 W/m·K. The inner surface of the base plate is subjected to uniform heat flux generated by the resistance heaters inside. When steady operating conditions are reached, the outer surface temperature of the plate is measured to be 112°C. Disregarding any heat loss through the upper part of the iron,

<u>Assumption</u>

Heat conduction is steady state and unidimensional  2. thermal conductivity is constant. Heat supplied is not in the plate

4. we disregard heat loss

Heat flux=heat/area

\alpha/A=800W/160*10^-4

with direction to the surface been in the x direction,

the mathematical expression will be

\frac{d^2T}{dx^2}=0..............1

and \frac{-kdT(0)}{dx} =q_{0}=5000W/m^2

from fourier law, for conductivity

T(L)=T2=112C

b. integrating equation 1 twice we have\dT/dx=c1

T(x)=C1x+C2

C1 and C2 are arbitrary constant

at x=0 the boundary conditions become

-kC1=qo

C1=-(qo/k)

at x=L          

=T(L)=C1L+C2=T2

C2=T2-cL1

C2=T2+qoL/k

Juxtaposing C1 and C2 into the general equation , we have

T(x)=-qo/k+T2+qoL/k=qo(L-k)/k+T2

50000*(0.006-x)/60+112

833.3(0.006-x)+112

c. inner surface plate temperature is

T(0)=833.33(0.006-0)+112 ( using the derivation in answer b)

117 deg C

6 0
3 years ago
For a bronze alloy, the stress at which plastic deformation begins is 297 MPa and the modulus of elasticity is 113 GPa. (a) What
Alenkinab [10]

Answer:

a) 93.852 kN

b) 128.043 mm

Explanation:

Stress is load over section:

σ = P / A

If plastic deformation begins with a stress of 297 MPa, the maximum load before plastic deformation will be:

P = σ * A

316 mm^2 = 3.16*10^-4

P = 297*10^6 * 3.16*10^-4 = 93852 N = 93.852 kN

The stiffness of the specimen is:

k = E * A / l

k = 113*10^9 * 3.16*10^-4 / 0.128 = 279 MN/m

Hooke's law:

x' = x0 * (1 + P/k)

x' = 0.128 * (1 + 93.852*10^3 / 279*10^6) = 0.128043 m = 128.043 mm

5 0
3 years ago
Are commonly made out of cast iron and connect directly to the engine
iren [92.7K]

Answer:

The engine block.

Explanation:

3 0
3 years ago
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