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inna [77]
3 years ago
11

A minor road intersects a major 4-lane divided road with a design speed of 55mph and a median width of 8 feet. The intersection

is controlled with a stop sign on the minor road. If the design vehicle is a passenger car, determine the minimum sight distance required on the major road that will allow a stopped vehicle on the minor road to safely turn left if the approach grade on the minor road is 3%.

Engineering
1 answer:
Vedmedyk [2.9K]3 years ago
8 0

Answer:

stopping sight diatnce is =158 meters

Explanation:

The Design Speed =55 MPH(Miles per Hour)

1 miles per hour =0.447 meter pe second

then 55MPH=55*0.447=24.58 meter per second

Consider recation time =2.5 second

Consider Co efficinet friction is 0.35

As per the Stopping sight distance based on the Breaking distance+ Lag distance

SSD=Vt+(V^2/(2gf))

SSD=24.58+(24.58^2/(2*9.81*0.35))

The total stopping sight distance is =150 meters

if we consider the Approaching garde on the minor road is 3%

SSD=Vt+(V^2/(2g(f-n/100))

SSD=24.58+(24.58^2/(2*9.81*(0.35-3/100))

then stopping sight diatnce is =158 meters

See attachment for workings

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Air-conditioners consume a significant amount of electrical energy in buildings. Split air conditioner is a unitary system where
nydimaria [60]

Answer:

Evaporator,Compressor,Condensor ,Expanding valve

Explanation:

Split air conditioning :

  Split air conditioning means that, condensor unit or some time called outdoor unit is split from evaporator.It means that evaporator and condensor are placed at some distance.

The four component of split air conditioning system are as follows

1.Evaporator

 It absorb heat from room and produces the cooling effect.

2.Compressor

 It compresses the refrigerant which exits from evaporator.

3.Condensor

 It rejects the heat and cool the evaporator.

4.Expanding valve

  It allows to refrigerant to cool up to evaporator pressure.

6 0
3 years ago
compressors, the gas is often cooled while being compressed to reduce the power consumed by the compressor. explain how cooling
ASHA 777 [7]

The amount of work done by steady flow devices varies with the particular gas volume. The kinetic energy of gas particles decreases during cooling.

When the gas is subjected to intermediate cooling during compression, the gas specific volume is reduced, which lowers the compressor's power consumption. Compression is less adiabatic and more isothermal because the compressed gas must be cooled between stages since compression produces heat. The system's thermodynamic cycle's cold sink temperature is lowered by cooling the compressor coils. By increasing the temperature difference between the heat source and the cold sink, this improves efficiency.

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8 0
1 year ago
A program is seeded with 30 faults. During testing, 21 faults are detected, 15 of which are seeded faults and 6 of which are ind
Vesna [10]

Answer:

Estimated number of indigenous faults remaining undetected is 6

Explanation:

The maximum likelihood estimate of indigenous faults is given by,

N_F=n_F\times \frac{N_S}{n_S} here,

n_F = the number of unseeded faults = 6

N_S = number of seeded faults = 30

n_s = number of seeded faults found = 15

So NF will be calculated as,

N_F=6\times \frac{30}{15}=12

And the estimate of faults remaining is  N_F-n_F = 12 - 6 = 6

8 0
3 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
A tool chest has 650 N weight that acts through the midpoint of the chest. The chest is supported by feet at A and rollers at B.
erica [24]

Answer:

the value of horizontal force P is 170.625 N

the value of horizontal force at P = 227.5 N is that the block moves to right and this motion is due to sliding.

Explanation:

The first diagram attached below shows the free body diagram of the tool chest when it is sliding.

Let start out by calculating the friction force

F_f= \mu N_2

where :

F_f = friction force

\mu = coefficient of friction

N_2 = normal friction

Given that:

\mu = 0.3

F_f = 0.3 N_2

Using the equation of equilibrium along horizontal direction.

\sum f_x = 0

P - F_f = 0

P = 0.3 N_2   ----- Equation (1)

To determine the moment about point B ; we have the expression

\sum M_B  = 0

0 = N_2*70-W*35-P*100

where;

P = horizontal force

N_2 = normal force at support A

W = self- weight of tool chest

Replacing W = 650 N

0 = N_2*70-650*35-100*P

P = \frac{70 N_2-22750}{100} ----- equation (2)

Replacing  \frac{70 N_2-22750}{100}  for P in equation (1)

\frac{70N_2 -22750}{100} =0.3 N_2

N_2 = \frac{22750}{40}N_2 = 568.75 \ N

Plugging the value of N_2 = 568.75 \ N in equation (2)

P = \frac{70(568.75)-22750}{100} \\ \\ P = \frac{39812.5-22750}{100}  \\ \\ P = \frac{17062.5}{100}

P =170.625 N

Thus; the value of horizontal force P is 170.625 N

b)  From the second diagram attached the free body diagram; the free body diagram of the tool chest when it is tipping about point A is also shown below:

Taking the moments about point A:

\sum M_A = 0

-(P × 100)+ (W×35) = 0

P = \frac{W*35}{100}

Replacing 650 N  for W

P = \frac{650*35}{100}

P = 227.5 N

Thus; the value of horizontal force P, when the tool chest tipping about point A is 227.5 N

We conclude that the motion will be impending for the lowest value when P = 170.625 N and when P= 227.5 N

However; the value of horizontal force at P = 227.5 N is that the block moves to right and this motion is due to sliding.

5 0
3 years ago
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