Ask question

Ask question

All categories

2 years ago

14

This tool used to check horizontal alignment

1 answer:

-BARSIC- [3]2 years ago

3 0

the answer is level bar

hope it do helps you

You might be interested in

Motion is defined as a change in an object's position when compared to other objects around it. Mary Ann was watching a slug cra

lara [203]

Answer:

Explanation:

Do it yourself bum

4 0

4 years ago

The A-36 steel drill shaft of an oil well extends 12 000 ft into the ground. Assuming that the pipe used to drill the well is su

sladkih [1.3K]

check photo solvecheck photo solve

3 0

3 years ago

Q.13 In order to produce maximum starting torque in a split-phase motor, how many degrees out of phase should the start- and run

Phantasy [73]

Answer:

D. 90 degrees.

Explanation:

Torque is a rotational force which moves an object in other direction. There should be 90 degrees out of phase to start, run winding currents with each other. Torque is produced by the rotational motion of an object. The angle of the object must be 90 degrees set in order to create torque.

3 0

3 years ago

g Consider a thin opaque, horizontal plate with an electrical heater on its backside. The front end is exposed to ambient air th

xxTIMURxx [149]

Answer:

The electrical power is 96.5 W/m^2

Explanation:

The energy balance is:

Ein-Eout=0

$qe+\alpha sGs+\alpha skyGsky-EEb(Ts)-qc=0$

if:

Gsky=oTsky^4

Eb=oTs^4

qc=h(Ts-Tα)

$\alpha s=\frac{\int\limits^\alpha _0 {\alpha l Gl} \, dl }{\int\limits^\alpha _0 {Gl} \, dl }$

$\alpha s=\frac{\int\limits^\alpha _0 {\alpha lEl(l,5800 } \, dl }{\int\limits^\alpha _0 {El(l,5800)} \, dl }$

if Gl≈El(l,5800)

$\alpha s=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))$

lt= 2*5800=11600 um-K, at this value, F=0.941

$\alpha s=(0.8*0.941)+0.3(1-0.941)=0.77$

The hemispherical emissivity is equal to:

$E=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))$

lt=2*333=666 K, at this value, F=0

$E=0+(1-0.7)(1)=0.3$

The hemispherical absorptivity is equal to:

$qe=EoTs^{4}+h(Ts-T\alpha )-\alpha sGs-\alpha oTsky^{4}=(0.3*5.67x10^{-8}*333^{4})+10(60-20)-(0.77-600)-(0.3*5.67x10^{-8}*233^{4})=96.5 W/m^{2}$

3 0

3 years ago

Some aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 35 MPa m . It has bee

GaryK [48]

Answer: Fracture will not occur since Kc (32.2 MPa√m) ∠ KIc (35 MPa√m).

Explanation:

in this question we are asked to determine if an aircraft will fracture for a given fracture toughness.

let us begin,

from the question we have that;

stress = 325 MPa

fracture toughness (KIc) = 35 MPa√m

the max internal crack length = 1.0 m

using the formula;

Y = KIc/σ√(πα) ---------------(1)

solving for Y we have;

Y = 35 (MPa√m) / 250 (MPa) √(π × 2×10⁻3/2m)

Y = 2.50

so to calculate the fracture roughness;

Kc = Y × σ√(πα) = 2.5 × 3.25√(π × 1×10⁻³/2) = 32.2 MPa√m

Kc = 32.2 MPa√m

From our results we can say that fracture will not occur since Kc (32.2 MPa√m) is less than KIc (35 MPa√m) of the material.

cheers i hope this helps!!!!

8 0

3 years ago