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uysha [10]
4 years ago
8

Technician A says that a low-restriction exhaust system could prevent a back pressure-type vacuum-controlled EGR valve from open

ing correctly. Technician B says restricted exhaust can cause the EGR valve position sensor to fail. Which technician is correct
Physics
1 answer:
Alborosie4 years ago
7 0

Answer:

Both technicians are correct.

Explanation:

Exhaust gas recirculation (EGR) contributes to limit the production of polluted gas mixture during internal engine combustion via an EGR valve. With the help of Exhaust gas recirculation (EGR), nitrogen oxides production is reduced drastically especially when the engine runs on diesel. Black smoke, engines making knocking sounds could be a clear indication of a bad EGR valve especially when the valve passage is blocked due to carbon deposits, or dirt from fuel blocking the valve opening thereby decreasing the power output, or in the case of the knock sounds, the fuel gets ignited early when the machine's operation at a given time is low. So, the Technicians are correct with the statements they made.

   

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A 0.100-kilogram apple falls from a height of 1.50 meters to 1.00 meters. Ignoring frictional effects, what is the kinetic energ
neonofarm [45]
Before the fall, at 1.50 m, all the energy is potential and none is kinetic.

As it's falling the apple has both kinetic and potential energies. The decrease in potential energy is equal to the increase in kinetic energy

KE = -ΔPE
= -mgΔh
= -(0.100 kg)(9.81 m/s²)(1.00 m - 1.50 m)
= 0.491 J
3 0
3 years ago
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The
Fofino [41]

Answer:

Wave Variables

In the chapter on motion in two dimensions, we defined the following variables to describe harmonic motion:

Amplitude—maximum displacement from the equilibrium position of an object oscillating around such equilibrium position

Frequency—number of events per unit of time

Period—time it takes to complete one oscillation

For waves, these variables have the same basic meaning. However, it is helpful to word the definitions in a more specific way that applies directly to waves:

].

Explanation:

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3 years ago
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An electron is in a vacuum near Earth's surface and located at y = 0 m on a vertical y axis. At what value of y should a group o
Bingel [31]

Answer:

the bunch of 23 electrons must be placed on y axis at coordinate y = - 24.35 m.

Explanation:

As we know that the gravitational force on electron at y = 0 is counter-balanced by the weight of the electron

So we have

\frac{kq_1q_2}{r^2} = mg

here we have

q_1 = e

q_2 = 23 e

m = 9.11 \times 10^{-31} kg

also we know that

e = 1.6 \times 10^{-19} C

so we will have

\frac{(9\times 10^9)(1.6 \times 10^{-19})(23\times 1.6 \times 10^{-19})}{r^2} = (9.11 \times 10^{-31})(9.81)

\frac{5.3 \times 10^{-27}}{r^2} = 8.94 \times 10^{-30}

r = 24.35 m

so the bunch of 23 electrons must be placed on y axis at coordinate y = - 24.35 m.

7 0
3 years ago
PLEASE HELP!!
astraxan [27]

Answer:

Explanation:

\lambda\\ = v/f

^That is the formula we are going to use.

Now, we were given the speed (v), which is 20.

Now we need to find frequency, in order to solve for the wavelength.

Frequency is the amount of waves in a fixed unit of one second, meaning our F value is the value of 5 divided by 4.

5/4 = 1.25

Therefore our F is 1.25

Now lets plug it in

\lambda\\ = v/f

\lambda\\ = 20/1.25

\lambda\\ = 16

Conversion:

\lambda\\ = 8

3 0
3 years ago
Human centrifuges are used to train military pilots and astronauts in preparation for high-g maneuvers. A trained, fit person we
jeka57 [31]

(a) 4.14 rad/s^2

The relationship beween centripetal acceleration and angular speed is

a=\omega^2 r

where

\omega is the angular speed

r is the radius of the circular path

Here we gave

a = 9g = 88.2 m/s^2 is the centripetal acceleration

r = 5.15 m is the radius

Solving for \omega, we find:

\omega = \sqrt{\frac{a}{r}}=\sqrt{\frac{88.2 m/s^2}{5.15 m}}=4.14 rad/s^2

(b) 21.3 m/s

The relationship between the linear speed and the angular speed is

v=\omega r

where

v is the linear speed

\omega is the angular speed

r is the radius of the circular path

In this problem we have

\omega=4.14 rad/s

r = 5.15 m

Solving the equation for v, we find

v=(4.14 rad/s)(5.15 m)=21.3 m/s

7 0
3 years ago
Read 2 more answers
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