All categories

3 years ago

PLEASE HURRY TIMED TEST

Engineering

2 answers:

ladessa [460]3 years ago

5 0

Answer:

Explanation:

Vladimir [108]3 years ago

3 0

EXHAUST VALVE OPENS....... 1...... AAAAAAA

You might be interested in

Steel balls 10 mm in diameter are annealed by heating to 1150 k and then slowly cooling to 450 k in an air environment for which

Darina [25.2K]

Answer:

The answer is below

Explanation:

Given that:

ρ = 800 kg/m3, c = 600 J/kg-K, k = 40 W/m-K, Initial temperature = Ti = 1150,

Environment temperature = T = 450 K, Final temperature = T∞ = 325 K

Diameter = 10 mm = 0.01 m, A = 6

The estimated time for cooling process (t) is given as:

$t=\frac{\rho dc}{hA} ln\frac{T_i-T_\infty}{T-T_infty}=\frac{7800*0.01*600}{25*6} ln\frac{1150-325}{450-325}\\ \\t=589\ s\\t=0.1635\ h$

The estimated cooling time is 589 s

4 0

3 years ago

Which term refers to the impurities found during the welding process ?

Aleks04 [339]

Answer:

idk

Explanation:

idk

6 0

4 years ago

What is 222 divided by 11.2

valentinak56 [21]

Answer:

The answer of this question is 19.89142857

3 0

3 years ago

Read 2 more answers

A 1.5-m-long aluminum rod must not stretch more than 1 mm and the normal stress must not exceed 40 MPa when the rod is subjected

Pavlova-9 [17]

Answer:

the required diameter of the rod is 9.77 mm

Explanation:

Given:

Length = 1.5 m

Tension(P) = 3 kN = 3 × 10³ N

Maximum allowable stress(S) = 40 MPa = 40 × 10⁶ Pa

E = 70 GPa = 70 × 10⁹ Pa

δ = 1 mm = 1 × 10⁻³ m

The required diameter(d) = ?

a) for stress

The stress equation is given by:

$S = \frac{P}{A}$

A is the area = πd²/4 = (3.14 × d²)/4

$S = \frac{P}{(\frac{3.14*d^{2} }{4}) }$

$S = \frac{4P}{{3.14*d^{2} } }$

$3.14*S*{d^{2}} = {4P}$

${d^{2}} =\frac{4P}{3.14*S}$

$d= \sqrt{\frac{4P}{3.14*S} }$

Substituting the values, we get

$d= \sqrt{\frac{4*3*10^{3} }{3.14*40*10^{6} } }$

$d= \sqrt{\frac{12000 }{125600000 } }$

$d= \sqrt{9.55*10^{-5} }$

d = (9.77 × 10⁻³) m

d = 9.77 mm

b) for deformation

δ = (P×L) / (A×E)

A = (P×L) / (E×δ) = (3000 × 1.5) / (1 × 10⁻³ × 70 × 10⁹) = 0.000063

d² = (4 × A) / π = (0.000063 × 4) / 3.14

d² = 0.0000819

d = 9.05 × 10⁻³ m = 9.05 mm

We use the larger value of diameter = 9.77 mm

3 0

3 years ago

An AC circuit has a resistor, capacitor and inductor in series with a 120 V, 60 Hz voltage source. The resistance of the resisto

aliina [53]

Answer:

(i) 3.5385 ohm, 3.768 ohm (ii) 39.89 A (III) 4773.857 W (vi) 348 var (vii) 0.9973 (viii) 4.1796°

Explanation:

We have given voltage V =120 volt

Frequency f=60 Hz

Resistance R =3 ohm

Inductance L =0.01 H

Capacitance C =0.00075 farad

(i) reactance of of inductor $X_L=\omega L=2\pi fL=2\times 3.14\times 60\times 0.01=3.768ohm$

$X_C=\frac{1}{\omega C}=\frac{1}{2\times 3.14\times 60\times 0.00075}=3.5385ohm$

(ii) Total impedance $Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{3^2+(3.768-3.5385)^2}=3.008ohm$

Current $i=\frac{V}{Z}=\frac{120}{3.008}=39.89A$

(viii) power factor $cos\Phi =\frac{R}{Z}=\frac{3}{3.008}=0.9973$

(VII) $cos\Phi =0.9973$

$\Phi =4.1796^{\circ}$

So power factor angle is 4.1796°

(iii) Apparent power $P=VICOS\Phi =120\times 39.89\times 0.9973=4773.875W$

(vi) Reactive power $Q=VISIN\Phi =120\times 39.89\times SIN4.17^{\circ}=348var$

5 0

3 years ago