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alekssr [168]
4 years ago
8

failure to yield the right-of-way to another vehicle or pedestrian is the primary Collision factor in about 20% of fatal and inj

ury collisions in California what are the keys to avoid accidents caused by right away violations​
Engineering
1 answer:
laila [671]4 years ago
8 0

You could just create a turn on green arrow. Have a button for pedestrians. The only way for them to turn is if they have a green arrow and the green arrow will only appear when pedestrians are stopped or finished walking and all cars are clear or fully stopped

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If a resistor only has three color bands, it has a tolerance of 20 percent. What would this mean for a resistor specified at 10,
Aleksandr-060686 [28]
The correct answer has to be c
8 0
2 years ago
A gear train has two gears. The driver gear has 8 teeth and a diametral pitch of 6 teeth/inch. the follower gear has 24 teeth. W
Sliva [168]

Answer:

18 teeth/inch

Explanation:

Given that: i. driver gear has 8 teeth and diametral pitch of 6 teeth/inch.

                  ii. follower gear has 24 teeth.

Let the followers diametral pitch be represented by x.

Then,

8 teeth ⇒  6 teeth/inch

24 teeth ⇒ x teeth/inch

So that;

x = \frac{24 x 6}{8}

   = \frac{144}{8}

   = 18 teeth/inch

The diametral is 18 teeth/inch

3 0
3 years ago
Process synchronization can be done on
marishachu [46]

Answer:

it is c) both (a) and (b)

Explanation:

5 0
3 years ago
In a Rankine cycle, superheated steam that enters the turbine at 1273.15 K and 1.8 MPa is then expanded to a vapor at 0.1 MPa. W
GrogVix [38]

Answer:

The shaft work generated per kilogram is -1.3 \frac{MJ}{kg}

Explanation:

Given:

Temperature T = 1273.15 K

Initial Pressure P_{1} = 1.8 MPa

Final pressure P_{2} = 0.1 MPa

From the table superheated,

h_{i} = 4635 \frac{K J}{Kg} and  h_{f} = 2706.54 \frac{K J}{Kg}

Work done by shaft is,

 W = h_{f} - h_{i}

 W = 2706.54 - 4635

 W = -1928.46 \frac{kJ}{kg}

But here efficiency is 0.56,

So work generated per kg is,

Work = 0.56 \times(- 1928.46)

Work = -1.3 \frac{MJ}{kg}

Therefore, the shaft work generated per kilogram is -1.3 \frac{MJ}{kg}

6 0
3 years ago
A 78-percent efficient 12-hp pump is pumping water from a lake to a nearby pool at a rate of 1.5 ft3/s through a constant-diamet
fenix001 [56]

Answer:

irreversible head loss is 38.51 ft

mechanical power 6.55 hp

Explanation:

Given data:

Pump Power 12 hp = 8948.39 watt = 6600 lbs ft/sec

Q = 1.5 ft^3/s

Pump actually delivers P'  = \eta P = 0.78 \times 8948.39 = 6979.74 watt

Power that water gains= mgh = \rho \phi gh = r \phi h

P = 62.4 \times 1.5 \times 32 = 2995.2 lbs ft/sec

hence Power lost = 6600 - 2995.2  3604.8 lbs ft/sec = 6.55 hp

head losshl = \frac{3604.8}{r \phi} = \frac{3604.8}{62.4 \times 1.5}

hl = 38.51 ft

8 0
3 years ago
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