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vovangra [49]
2 years ago
14

9. An automobile's oxygen sensor output needs

Engineering
1 answer:
Maslowich2 years ago
5 0

Answer:

the answer is Letter C

Explanation:

DON'T TRULY TRUST ME PLEASE..

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Tech A says that both OSHA and the EPA can inspect facilities for violations. Tech B says that a shop safety rule does not have
balu736 [363]

Answer:

A is right, depending on what you mean the health organization OSHA is aloud to search faculty if they can or may be affecting peoples health. With a proper search it can be made clear if its safe or not and the OSHA is aloud to do it without your consent to it.

8 0
3 years ago
Two reversible cycles operate between hot and cold reservoirs at temperature TH and TC, respectively. If one is a power cycle an
Reika [66]

Answer:

COP_{HP} = \frac{1}{\eta_{th}}

Explanation:

The coefficient of performance of the reversible heat pump is:

COP_{HP} = \frac{T_{H}}{T_{H}-T_{C}}

COP_{HP} = \frac{1}{1-\frac{T_{C}}{T_{H}} }

The thermal efficiency of the reversible power cycle is:

\eta_{th} = 1 - \frac{T_{C}}{T_{H}}

After a quick comparison between both expressions, the following relation is found:

COP_{HP} = \frac{1}{\eta_{th}}

5 0
4 years ago
Are we living in a simulation! <br> True <br> False
olchik [2.2K]
False

i would hope:):):)
4 0
3 years ago
Read 2 more answers
A light train made up of two cars is traveling at 90 km/h when the brakes are applied to both cars. Know that car A has a mass o
posledela

Answer:

a) d=236.280\,m, b) F_{coupling} = -8848\,N The real force has the opposite direction.

Explanation:

a) Let assume that train moves on the horizontal ground. An equation for the distance travelled by the train is modelled after the Principle of Energy Conservation and Work-Energy Theorem:

K_{A} = W_{brake}

\frac{1}{2}\cdot m_{train} \cdot v^{2} = F_{brakes}\cdot d

d = \frac{m_{train}\cdot v^{2}}{2\cdot F_{brakes}}

d = \frac{(51000\,kg)\cdot [(90\,\frac{km}{h} )\cdot (\frac{1000\,m}{1\,km} )\cdot (\frac{1\,h}{3600\,s} )]^{2}}{2\cdot (82000\,N)}

d=194.360\,m

b) The acceleration experimented by both trains are:

a = -\frac{v_{o}^{2}}{2\cdot d}

a = -\frac{[(90\,\frac{km}{h} )\cdot (\frac{1000\,m}{1\,km} )\cdot (\frac{1\,h}{3600\,s})]^{2}}{2\cdot (194.360\,m)}

a = -1.608\,\frac{m}{s^{2}}

The coupling force in the car A can derived of the following equation of equilibrium:

\Sigma F = F_{coupling} - F_{brakes} = m_{A}\cdot a

The coupling force between cars is:

F_{coupling} = m_{A}\cdot a + F_{brakes}

F_{coupling} = (31000\,kg)\cdot(-1.608\,\frac{m}{s^{2}} )+41000\,N

F_{coupling} = -8848\,N

The real force has the opposite direction.

8 0
3 years ago
A cubic shaped box has a side length of 1.0 ft and a mass of 10 lbm is sliding on a frictionless horizontal surface towards a 30
natulia [17]

Explanation  & answer:

Assuming a smooth transition so that there is no abrupt change in slopes to avoid frictional loss nor toppling, we can use energy considerations.

Initially, the cube has a kinetic energy of

KE = mv^2/2 = 10 lbm * 20^2 ft^2/s^2  / 2 = 2000 lbm-ft^2 / s^2

At the highest point when the block stops, the gain in potential energy is

PE = mgh = 10 lbm * 32.2 ft/s^2 * h ft = 322 lbm ft^2/s^2

By assumption, there was no loss in energies, we equate PE = KE

322h lbm ft^2/s^2 = 2000 lbm ft^2/s^2

=>

h = 2000 /322 = 6.211 (ft)

distance up incline = h / sin(30) = 12.4 ft

6 0
3 years ago
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