The minimum safe working distance from exposed electrical conductors

4. Two technicians are discussing the evaporative emission monitor. Technician A says that serious monitor faults cause a blinki

snow_lady [41]

Answer:

The correct option is;

Neither Technician A nor B

Explanation:

The evaporative emission monitor or Evaporaive Emission Control System EVAP System monitors enables the Power Control Module of the car to check fuel system leak integrity and the vapor consumption efficiency during engine combustion

It is a requirement of EPA on cars to check the emission of smug forming evaporates from cars

Serious monitor faults can cause the turning on of the check engine lights and the vehicle will not pass OBD II test, but it will not lead to engine shutdown

It runs when the engine is 15 to 85% full and the TP sensor is between 9% and 35%.

Therefore, the correct option is that neither Technician A nor B are correct.

3 0

3 years ago

The acceleration (in m/s^2) of a linear slider (undergoing rectilinear motion) within a If the machine can be expressed in terms

Inga [223]

Answer:

47.91 sec

Explanation:

it is given that $\alpha =\frac{1}{4v^{2}}$

at t=0 velocity =0 ( as it is given that it is starting from rest )

we have to find time at which velocity will be 3.3 $\frac{m}{sec^{2}}$

we know that $\alpha =\frac{dv}{dt}=\frac{1}{4v^{2}}$

$4v^{2}dv=dt$

integrating both side

$\frac{4v^{3}}{3}=t+c$ ---------------eqn 1

at t=o it is given that v=0 putting these value in eqn 1 c=0

so $\frac{4v^{3}}{3}=t$

when v= 3.3 $\frac{m}{sec^{2}}$

t= $\frac{4}{3}\times 3.3^{3}$

=47.91 sec

6 0

3 years ago

In a surface grinding operation, the wheel diameter = 8.0 in, wheel width = 1.0 in, wheel speed = 6000 ft/min, work speed = 40 f

Katen [24]

Answer: the answer will be d because it is the right one to be

Explanation:

7 0

3 years ago

Plant scientists would not do which of the following?

zavuch27 [327]

Explanation:

i think option 4 is correct answer because itsrelated to animal not plants.

6 0

3 years ago

Read 2 more answers

At a certain location, wind is blowing steadily at 5 mph. Suppose that the mass density of air is 0.0796 lbm/ft3 and determine t

nlexa [21]

Answer:

The radius of a wind turbine is 691.1 ft

The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m

Explanation:

Given;

power generation potential (PGP) = 1000 kW

Wind speed = 5 mph = 2.2352 m/s

Density of air = 0.0796 lbm/ft³ = 1.275 kg/m³

Radius of the wind turbine r = ?

Wind energy per unit mass of air, e = E/m = 0.5 v² = (0.5)(2.2352)²

Wind energy per unit mass of air = 2.517 J/kg

PGP = mass flow rate * energy per unit mass

PGP = ρ*A*V*e

$PGP = \rho *\frac{\pi r^2}{2} *V*e \\\\r^2 = \frac{2*PGP}{\rho*\pi *V*e} , r=\sqrt{ \frac{2*PGP}{\rho*\pi *V*e}} = \sqrt{ \frac{2*10^6}{1.275*\pi *2.235*2.517}}$

r = 210.64 m = 691.1 ft

Thus, the radius of a wind turbine is 691.1 ft

PGP = CVᵃ

For best design of wind turbine Betz limit (c) is taken between (0.35 - 0.45)

Let C = 0.4

PGP = Cvᵃ

take log of both sides

ln(PGP) = a*ln(CV)

a = ln(PGP)/ln(CV)

a = ln(1000)/ln(0.4 *2.2352) = 7.73

The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m

5 0

2 years ago