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IgorLugansk [536]
2 years ago
6

How do traffic lights involve science??

Engineering
2 answers:
satela [25.4K]2 years ago
6 0

Answer:

Many would pair the engineering and science behind it together. The sensor requires have a detective system for when cars pass by. It can also cycle the lights (change from red, green, and yellow). Traffic can flow uninterrupted on the main road. So, science is an important part.

GaryK [48]2 years ago
4 0
When a vehicle on a side road arrives at the intersection, a sensor will detect it and cycle the lights to allow traffic on the side road to pass through. In this way, traffic can flow uninterrupted on the main road unless and until traffic on a side road appears.
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g a heat engine is located between thermal reservoirs at 400k and 1600k. the heat engine operates with an efficiency that is 70%
Bond [772]

Answer:

<em>Heat rejected to cold body = 3.81 kJ</em>

Explanation:

Temperature of hot thermal reservoir Th = 1600 K

Temperature of cold thermal reservoir Tc = 400 K

<em>efficiency of the Carnot's engine = 1 - </em>\frac{Tc}{Th}<em> </em>

eff. of the Carnot's engine = 1 - \frac{400}{600}

eff = 1 - 0.25 = 0.75

<em>efficiency of the heat engine = 70% of 0.75 = 0.525</em>

work done by heat engine = 2 kJ

<em>eff. of heat engine is gotten as = W/Q</em>

where W = work done by heat engine

Q = heat rejected by heat engine to lower temperature reservoir

from the equation, we can derive that

heat rejected Q = W/eff = 2/0.525 = <em>3.81 kJ</em>

6 0
2 years ago
What s the difference between buying a modded account or buying a booster for gta online
lorasvet [3.4K]

Answer:

The differences are the benefits, what outweighs the other. That is the difference.

8 0
2 years ago
A. An 80-percent efficient pump with a power input of 20 hp is pumping water from a lake to a nearby pool at ar ate of 1.5 ft3 /
ss7ja [257]

Answer:

A) mechanical power used to overcome frictional effects in piping = 2.373 HP

B) efficiency = 66.82%

Pressure difference = 147 KPa

Explanation:

We are given;

efficiency of pump; η = 80% = 0.8

power input of pump;P_pump,in = 20 hp

rate being pumped; Q = 1.5 ft³/s = 0.04248 m³/s

free surface of pool; Δz = 80 ft = 24.384 m

mechanical pumping power used to deliver water is;

W_u = η × P_pump,in

W_u = 0.8 × 20

W_u = 16 Hp

Now, change in total mechanical energy of water will be equal to the change in potential energy.

Thus, it is expressed as;

ΔE_mech = ρ × Q × g × Δz

Where ρ is density of water = 1000 kg/m³

Thus;

ΔE_mech = 1000 × 0.04248 × 9.81 × 24.384 = 10,161.5150592 J/s

Converting to HP gives;

ΔE_mech = 13.627 HP

Now, mechanical power used to overcome frictional effects is given by;

W_frict = W_u - ΔE_mech

W_frict = 16 - 13.627

W_frict = 2.373 HP

B) We are now given;

W_elect,in = 15.4 KW

Volumetric flow rate; q = 70 l/s = 0.07 m³/s

Height; h = 15 m

Mass flow rate; m' = qρ

Where ρ is density of water = 1000 kg/m³.

m' = 0.07 × 1000

m' = 70 kg/s

Now,

ΔE_mech = workdone through height of 15m

Thus;

ΔE_mech = m'gh

ΔE_mech = 70 × 9.8 × 15

ΔE_mech = 10,290 W = 10.29 Kw

efficiency of the pump–motor unit = ΔE_mech/(W_elect,in) = 10.29/15.4 = 0.6682 or 66.82%

Pressure difference = ρgh = 1000 × 9.8 × 15 = 147,000 Pa = 147 KPa

3 0
2 years ago
My brother john has never learned this lesson is that a sentence
oee [108]

Answer:

Yes

Explanation:

You just need to add a period behind the word "lesson", and make it My brother john has never learned this lesson.

6 0
2 years ago
A rear wheel drive car of mass 1000 kg is accelerating with a constant acceleration without slipping from 0 to 60 m/s in 1 min.
tester [92]

Answer:

500 N

Explanation:

Given;

Mass of the car, M = 1000 kg

initial speed of the car, u = 0 m/s

Final speed of the car, v = 60 m/s

Time, t = 1 min = 60 s

Now,

Force, F is given as:

F = Ma

where,

a is the acceleration

From the Newton's equation of motion, we have

v = u + at

on substituting the values, we get

60 = 0 + a × 60

or

a = 1 m/s²

Thus,

Force = 1000 × 1 = 1000 N

now,

this force will be equal to the friction force provided by the rear wheels

let the friction force on a single rear wheel be 'f'

thus,

2f = 1000 N

or

f = 500 N

5 0
3 years ago
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