All categories

2 years ago

Engineers are problem blank who use critical thinking to create new solutions.

Engineering

2 answers:

Nana76 [90]2 years ago

7 0

Answer:

problem solvers

Explanation:

irakobra [83]2 years ago

7 0

Correct I think rare

You might be interested in

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

Vlad [161]

Answer:

Explanation:

you need more information x

7 0

3 years ago

At an impaired driver checkpoint, the time required to conduct the impairment test varies (according to an exponential distribut

professor190 [17]

Answer:

Option (d) 2 min/veh

Explanation:

Data provided in the question:

Average time required = 60 seconds

Therefore,

The maximum capacity that can be accommodated on the system, μ = 60 veh/hr

Average Arrival rate, λ = 30 vehicles per hour

Now,

The average time spent by the vehicle is given as

⇒ $\frac{1}{\mu(1-\frac{\lambda}{\mu})}$

thus,

on substituting the respective values, we get

Average time spent by the vehicle = $\frac{1}{60(1-\frac{30}{60})}$

Average time spent by the vehicle = $\frac{1}{60(1-0.5)}$

Average time spent by the vehicle = $\frac{1}{60(0.5)}$

Average time spent by the vehicle = $\frac{1}{30}$ hr/veh

Average time spent by the vehicle = $\frac{1}{30}\times60$ min/veh

[ 1 hour = 60 minutes]

thus,

Average time spent by the vehicle = 2 min/veh

Hence,

Option (d) 2 min/veh

7 0

3 years ago

Water flows through a pipe at an average temperature of T[infinity] = 70°C. The inner and outer radii of the pipe are r1 = 6 cm

Paul [167]

Answer:

The differential equation and the boundary conditions are;

A) -kdT(r1)/dr = h[T∞ - T(r1)]

B) -kdT(r2)/dr = q'_s = 734.56 W/m²

Explanation:

We are given;

T∞ = 70°C.

Inner radii pipe; r1 = 6cm = 0.06 m

Outer radii of pipe;r2 = 6.5cm=0.065 m

Electrical heat power; Q'_s = 300 W

Since power is 300 W per metre length, then; L = 1 m

Now, to the heat flux at the surface of the wire is given by the formula;

q'_s = Q'_s/A

Where A is area = 2πrL

We'll use r2 = 0.065 m

A = 2π(0.065) × 1 = 0.13π

Thus;

q'_s = 300/0.13π

q'_s = 734.56 W/m²

The differential equation and the boundary conditions are;

A) -kdT(r1)/dr = h[T∞ - T(r1)]

B) -kdT(r2)/dr = q'_s = 734.56 W/m²

6 0

3 years ago

A fatigue test was conducted in which the mean stress was 90 MPa (13050 psi), and the stress amplitude was 190 MPa (27560 psi).

Gwar [14]

Answer:

a) 280MPa

b) -100MPa

c) -0.35

d) 380 MPa

Explanation:

GIVEN DATA:

mean stress $\sigma_m = 90MPa$

stress amplitude $\sigma_a = 190MPa$

a) $\sigma_m =\frac{\sigma_max+\sigma_min}{2}$

$90 =\frac{\sigma_{max}+\sigma_{min}}{2}$ --------------1

$\sigma_a =\frac{\sigma_{max}-\sigma_{min}}{2}$

$190 = \frac{\sigma_{max}-\sigma_{min}}{2}$ -----------2

solving 1 and 2 equation we get

$\sigma_{max} = 280MPa$

b) $\sigma_{min} = - 100MPa$

stress ratio $=\frac{\sigma_{min}}{\sigma_{max}}$

$=\frac{-100}{280} = -0.35$

d)magnitude of stress range

$=(\sigma_{max} -\sigma_{min})$

= 280 -(-100) = 380 MPa

3 0

3 years ago

A 3-phase , 1MVA, 13.8kV/4160V, 60 Hz, transformer with Y-Delta winding connection is supplying a3-phase, 0.75 p.u. load on the

Tanya [424]

Answer:

a) 23.89 < -25.84 Ω

b) 31.38 < 25.84 A

c) 0.9323 leading

Explanation:

A) Calculate the load Impedance

current on load side = 0.75 p.u

power factor angle = 25.84

$I_{load}$ = 0.75 < 25.84°

attached below is the remaining part of the solution

B) Find the input current on the primary side in real units

load current in primary = 31.38 < 25.84 A

C) find the input power factor

power factor = 0.9323 leading

attached below is the detailed solution

8 0

3 years ago