After adjusting your seat, your___ should be as closest possible to the back rest.

When does the vc-turbo engine use lower compression ratios?.

Veronika [31]

Answer:

Explanation:

The VC-T engine (for "variable compression, turbocharged") can adjust its compression ratio between 8:1 and 14:1 on the fly, offering high-compression efficiency under light loads and the low compression needed for turbocharged power under hard acceleration.

7 0

2 years ago

All air-conditioning units must be grounded electrically to

Tpy6a [65]

Answer:

Prevent electrical shock

Grounding is a non current carrying conductor mainly used to guard against hazards due to leakage in electric circuits

Explanation:

The grounding refers to the connection of an electrical equipment exposed metallic parts to the ground to serve as a source of current flow in the event of an insulation failure will cause the fuses to trip thereby isolating or removing electric power from the device

Grounding also prevents the accumulation of static electricity which can be a source of fire in inflammable areas.

8 0

3 years ago

The terms batten seam, standing seam, and flat seam all describe types of:

abruzzese [7]

Answer:

 (A) architectural sheet metal roofing

Explanation:

By the name itself we can judge that the 'Architectural sheet metal roofing' is a kind of metal roofing.

And these type of metal roofing is primarily used for small and big houses, small buildings and as well as in a building that is for commercial use they can be totally flat as well as little bit sloped.

And the words similarly like batten and standing seam, and flat seam all tells us that these are the types of architectural sheet metal roofing.

5 0

3 years ago

Explain three examples of workshop

stealth61 [152]

Answer:

Invitational Workshop

An invitational workshop is what many of us know. It’s what Lucy Calkins has made famous through the Reading and Writing Workshop. In the invitational workshop, the instructor usually hosts a minilesson. This minilesson is intended to meet the needs of the majority of children in the classroom. Afterward, the children are “invited” to employ the skills or strategy for the minilesson during workshop time, where students work independently or in small groups

Explanation:

7 0

2 years ago

Read 2 more answers

ASAE 1060 Steel wire (1 mm diameter) is coated with copper to form a composite with a diameter of 2mm. Use the following propert

k0ka [10]

Answer:

a) $E_{m}$ = 133.75 Gpa

b) Fnet = 560 N

c) thermal expansion of the composite material = 14.31 $10^{-6 }$ / °C

Explanation:

Solution:

a) Elastic Modulus of the composite:

Area of steel wire = $\frac{\pi }{4}$ x ( $0.001^{2}$ ) = 0.8 x $10^{-6}$ $m^{2}$

Area of Copper wire = $\frac{\pi }{4}$ x ( $0.002^{2}$ ) - 0.8 x $10^{-6}$ $m^{2}$

Area of Copper wire = 2.4 x $10^{-6}$ $m^{2}$

Young's Modulus of Composite mixture:

$E_{m}$ = $F_{st}$ $E_{st}$ + $F_{Cu}$ $E_{Cu}$ Equation 1

here,

$F_{st}$ = Stress in Steel

$F_{Cu}$ = Stress in Copper.

We know that,

F = P/A

F is inversely proportional to Area, so if area is large, stress will less and vice versa. So, Take

Ratio for area of steel = $\frac{0.8. 10^{-6} }{(0.8 + 2.4) .10^{-6} }$

Ratio for area of steel = $\frac{0.8}{3.2 }$ = 0.25

Similarly, for Copper,

Ratio for area of copper = $\frac{2.4. 10^{-6} }{(0.8 + 2.4) .10^{-6} }$

Ratio for area of copper = $\frac{2.4 }{3.2}$ = 0.75

Put these values in equation 1:

$E_{m}$ = $F_{st}$ $E_{st}$ + $F_{Cu}$ $E_{Cu}$

$E_{m}$ = (0.25) $E_{st}$ + (0.75) $E_{Cu}$

We are given that,

$E_{st}$ = 205 Gpa

$E_{Cu}$ = 110 Gpa

So,

$E_{m}$ = (0.25) (205 Gpa) + (0.75) (110 GPa)

$E_{m}$ = 51.25GPa + 82.5 Gpa

Hence, the Elastic Modulus of the composite will be:

$E_{m}$ = 133.75 Gpa

b) maximum force:

Fnet = Fst + Fcu

We know that F = (Yield Stress x Area)

F = fst x Ast + fcu x Acu

And we are given that,

Yield stress of Steel = 280 Mpa

Yield stress of Copper = 140 Mpa

And,

Ast = 0.8 x $10^{-6}$ $m^{2}$

Acu = 2.4 x $10^{-6}$ $m^{2}$

Just plugging in the values, we get:

F = (280 Mpa) (0.8 x $10^{-6}$ $m^{2}$ ) + (140 Mpa) (2.4 x $10^{-6}$ $m^{2}$ )

F = 224 + 336

Fnet = 560 N ( because Mpa = $10^{6}$ N/ $m^{2}$ )

So, it means the composite will carry the maximum force of 560N

c) Coefficient of Thermal Expansion:

Strain on both material is same upon loading so,

(ΔL/L)st = (ΔL/L)cu

by thermal expansion equation:

( $\alpha .$ ΔT + $\frac{F}{A}$ $. \frac{1}{Est}$ ) = $\alpha .$ ΔT + $\frac{F}{A}$ $. \frac{1}{Ecu}$ )

Where $\alpha$ = Coefficient of Thermal expansion

Here, fst = -fcu = F

and ΔT = 1°

So,

Plugging in the values, we get.

( 10 x $10^{-6}$ x (1) + $\frac{F}{0.8.10^{-6} } . \frac{1}{205 . 10^{9} }$ ) = ( 17 x $10^{-6}$ x (1) + $\frac{-F}{2.4.10^{-6} } . \frac{1}{110 . 10^{9} }$ )

Solving for F, we get:

F = 0.71 N

Here,

fst = F = 0.71 N (Tension on Heating)

fcu = -F = 0.71 N ( Compression on Heating )

So, the combined thermal expansion of the composite material will be:

(ΔL/L)cu = ( 17 x $10^{-6}$ x (1°) + $\frac{-0.71}{2.4.10^{-6} } . \frac{1}{110 . 10^{9} }$ )

(ΔL/L)cu = ( 17 x $10^{-6}$ x (1°) - 2.69 x $10^{-6}$

combined thermal expansion of the composite material = 14.31 $10^{-6 }$ / °C

4 0

3 years ago