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3 years ago

What is the general term for a substance that is harmful to life and may cause death?

Engineering

1 answer:

lukranit [14]3 years ago

7 0

Toxic is the answer, toxic meaning may cause death or other health problems.

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Technician A says that a fully charged battery is less likely to freeze than a discharged battery. Technician B says that the st

zheka24 [161]

Answer : Technician B is correct

8 0

2 years ago

The roof of a refrigerated truck compartment consists of a layer of foamed urethane insulation (t2 = 21 mm, ki = 0.026 W/m K) be

lakkis [162]

Answer:

Tso = 28.15°C

Explanation:

given data

t2 = 21 mm

ki = 0.026 W/m K

t1 = 9 mm

kp = 180 W/m K

length of the roof is L = 13 m

net solar radiation into the roof = 107 W/m²

temperature of the inner surface Ts,i = -4°C

air temperature is T[infinity] = 29°C

convective heat transfer coefficient h = 47 W/m² K

solution

As when energy on the outer surface at roof of a refrigerated truck that is balance as

Q = $\frac{T \infty - T si }{\frac{1}{hA}+\frac{t1}{AKp}+\frac{t2}{AKi}+\frac{t1}{aKp}}$ .....................1

Q = $\frac{T \infty - Tso}{\frac{1}{hA}}$ .....................2

now we compare both equation 1 and 2 and put here value

$\frac{29-(-4)}{\frac{1}{47}+\frac{2\times0.009}{180}+\frac{0.021}{0.026}} = \frac{29-Tso}{\frac{1}{47}}$

solve it and we get

Tso = 28.153113

so Tso = 28.15°C

3 0

2 years ago

Ben leads a team of a few engineers at a robotics firm. A couple of them would like to improve their skills by taking additional

Anit [1.1K]

Answer:

i dont know

Explanation:

4 0

2 years ago

A three-phase wye-connected synchronous generator supplies a network through a transmission line. The network can absorb or deli

Amanda [17]

Answer:

the graph and the answer can be found in the explanation section

Explanation:

Given:

Network rated voltage = 24 kV

Impedance of network = 0.07 + j0.5 Ω/mi, 8 mi

Rn = 0.07 * 8 = 0.56 Ω

Xn = 0.5 * 8 = 4 Ω

If the alternator terminal voltage is equal to network rated voltage will have

Vt = 24 kV/√3 = 13.85 kV/phase

The alternative current is

$I_{a} =\frac{40x10^{6} }{\sqrt{3}*24x10^{3} } =926.2A$

$X_{s} =0.85\frac{13.85}{926.2} =12.7ohm$

The impedance Zn is

$\sqrt{0.56^{2}+4^{2} } =4.03ohm$

The voltage drop is

$I_{a} *Z_{n} =926.2*4.03=3732.58V$

$r_{dc} =\frac{voltage}{2*current} =\frac{13.85}{2*926.2} =7.476ohm$

rac = 1.2rdc = 1.2 * 7.476 = 8.97 Ω

The effective armature resistance is

$Z_{s} =\sqrt{R_{a}^{2}+X_{s}^{2} } =\sqrt{8.97^{2}+12.7^{2} } =15.55ohm$

The induced voltage for leading power factor is

$E_{F} ^{2} =OB^{2} +(BC-CD)^{2}$

if cosθ = 0.5

$E_{F} =\sqrt{(13850*0.5)^{2}+(\frac{3741}{2}-926.2*12.7)^{2} } =11937.51V$

if cosθ= 0.6

EF = 12790.8 V

if cosθ = 0.7

EF = 13731.05 V

if cosθ = 0.8

EF = 14741.6 V

if cosθ = 0.9

EF = 15809.02 V

if cosθ = 1

EF = 13975.6 V

The voltage regulation is

$\frac{E_{F}-V_{t} }{V_{t} } *100$

For each value:

if cosθ = 0.5

voltage regulation = -13.8%

if cosθ = 0.6

voltage regulation = -7.6%

if cosθ = 0.7

voltage regulation = -0.85%

if cosθ = 0.8

voltage regulation = 6.4%

if cosθ = 0.9

voltage regulation = 14%

if cosθ = 1

voltage regulation = 0.9%

the graph is shown in the attached image

for 10% of regulation the power factor is 0.81

8 0

3 years ago

Given the vector current density J = 10rho2zarho − 4rho cos2 φ aφ mA/m2:

Xelga [282]

Answer:

(a) Current density at P is $J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\$ .

(b) Total current I is 3.257 A

Explanation:

Because question includes symbols and formulas it can be misunderstood. In the question current density is given as below;

$J=10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}\\$

where $\textbf{a}_{\rho}$ and $\textbf{a}_{\phi}$ unit vectors.

(a) In order to find the current density at a specific point <em>(P)</em>, we can simply replace the coordinates in the current density equation. Therefore

$J(P(\rho=3, \phi=30^o,z=2))=10.3^2.2.\textbf{a}_{\rho}-4.3.(\cos(30^o)^2).\textbf{a}_{\phi}\\\\J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\$

(b) Total current flowing outward can be calculated by using the relation,

$I=\int {\textbf{J} \, \textbf{ds}$

where integral is calculated through the circular band given in the question. We can write the integral as below,

$I=\int\{(10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}).(\rho.d\phi.dz.\textbf{a}_{\rho}})\}\\\\I=\int\{(10\rho^2z).(\rho.d\phi.dz)\}\\\\\\$

due to unit vector multiplication. Then,

$I=10\int\(\rho^3z.dz.d\phi$

where $\rho=3,\ 0$ . Therefore

$I=10.3^3\int_2^{2.8}\(zdz.\int_0^{2\pi}d\phi\\I=270(\frac{2.8^2}{2}-\frac{2^2}{2} )(2\pi-0)=3257.2\ mA\\I=3.257\ A$

4 0

3 years ago