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Natasha2012 [34]
1 year ago
9

if this light stays on after the initial startup of the vehicle it signals a warning that there is something wrong with the serv

ice brake system. a) vehicle function indicators b) dashboard warning devices c) oil pressure indicator d) coolant temperature indicator e) electrical system indicator f) service brake system indicator
Engineering
1 answer:
NemiM [27]1 year ago
7 0

Service brake system indicator is the warning that there is something wrong with the service brake system.  Hence option f is correct.

<h3>What is indicator?</h3>

Amber-colored indicator lights can be found at the front, back, and occasionally on the left and right sides of the vehicle. Whether you're turning left, right, or into oncoming traffic, you use your indicators to signal your planned change of direction.

When this light turns on, one of two things will happen. Either the parking brake is engaged or the hydraulic fluid (brake fluid) in the master cylinder is low. Your brakes are made up of a system of hydraulic oil-filled tubes called brake lines.

Thus, service brake system indicator is the warning that there is something wrong with the service brake system.  Hence option f is correct.

To learn more about indicator, refer to the link below:

brainly.com/question/28093573

#SPJ1

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David wants to determine what his organization should do in the future, and set project targets. Which management function can D
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Answer:

Forecasting

Explanation:

Organizational forecasting is estimating of future events for the purpose of effective planning and decision making. This is one of the critical organizational functions as the forecast help managers to anticipate the future and to plan accordingly. Examples are financial forecasting, reporting, and operational metrics tracking, analyze financial data, create financial models use to predict future revenues.

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State two reasons why industrial files must be always be used fitted upon the individual
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3 years ago
A ramp from an expressway with a design speed of 30 mi/h connects with a local road, forming a T intersection. An additional lan
hram777 [196]

Answer:

the width of the turning roadway = 15 ft

Explanation:

Given that:

A ramp from an expressway with a design speed(u) =  30 mi/h connects with a local road

Using 0.08 for superelevation(e)

The minimum radius of the curve on the road can be determined by using the expression:

R = \dfrac{u^2}{15(e+f_s)}

where;

R= radius

f_s = coefficient of friction

From the tables of coefficient of friction for a design speed at 30 mi/h ;

f_s = 0.20

So;

R = \dfrac{30^2}{15(0.08+0.20)}

R = \dfrac{900}{15(0.28)}

R = \dfrac{900}{4.2}

R = 214.29 ft

R ≅ 215 ft

However; given that :

The turning roadway has stabilized shoulders on both sides and will provide for a onelane, one-way operation with no provision for passing a stalled vehicle.

From the tables of "Design widths of pavement for turning roads"

For a One-way operation with no provision for passing a stalled vehicle; this criteria falls under Case 1 operation

Similarly; we are told that the design vehicle is a single-unit truck; so therefore , it falls under traffic condition B.

As such in Case 1 operation that falls under traffic condition B  in accordance with the Design widths of pavement for turning roads;

If the radius = 215 ft; the value for the width of the turning roadway for this conditions = 15ft

Hence; the width of the turning roadway = 15 ft

5 0
3 years ago
Sea water with a density of 1025 kg/m3 flows steadily through a pump at 0.21 m3 /s. The pump inlet is 0.25 m in diameter. At the
myrzilka [38]

Answer:

\dot W_{pump} = 16264.922\,W\,(16.265\,kW)

Explanation:

The pump is modelled after applying Principle of Energy Conservation, whose form is:

\frac{P_{1}}{\rho\cdot g}+ \frac{v_{1}^{2}}{2\cdot g} +z_{1} + h_{pump}=\frac{P_{2}}{\rho\cdot g}+ \frac{v_{2}^{2}}{2\cdot g} +z_{2}

The head associated with the pump is cleared:

h_{pump} = \frac{P_{2}-P_{1}}{\rho\cdot g}+\frac{v_{2}^{2}-v_{1}^{2}}{2\cdot g}+(z_{2}-z_{1})

Inlet and outlet velocities are found:

v_{1} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.25\,m)^{2} }

v_{1} \approx 4.278\,\frac{m}{s}

v_{2} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.152\,m)^{2} }

v_{2} \approx 11.573\,\frac{m}{s}

Now, the head associated with the pump is finally computed:

h_{pump} = \frac{175\,kPa-81.326\,kPa}{(1025\,\frac{kg}{m^{3}} )\cdot (9.807\,\frac{m}{s^{2}} )} +\frac{(11.573\,\frac{m}{s} )^{2}-(4.278\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )} + 1.8\,m

h_{pump} = 7.705\,m

The power that pump adds to the fluid is:

\dot W_{pump} = \dot V \cdot \rho \cdot g \cdot h_{pump}

\dot W_{pump} = (0.21\,m^{3})\cdot (1025\,\frac{kg}{m^{3}})\cdot (9.807\,\frac{m}{s^{2}})\cdot(7.705\,m)

\dot W_{pump} = 16264.922\,W\,(16.265\,kW)

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3 years ago
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Answer:

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