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gulaghasi [49]
3 years ago
11

When the ignition is first turned on a click noise is heard from under the hood of a vehicle equipped with ETC. Technician A say

s that this is the normal operation of the ETC​ self-test. Technician B says that the TAC motor should not move unless the engine is running. Which technician is​ correct?
Physics
1 answer:
Ierofanga [76]3 years ago
4 0

Answer:

Technician A says that this is the normal operation of the ETC self -test is the correct answer.

Explanation:

An engine control unit (ECU), also widely referred to as an engine control module (ECM), is a type of electronic control device that controls an internal combustion engine with a series of actuators to ensure maximum engine performance.

It achieves so by reading values from a multitude of sensors within the engine bay, translating data using multidimensional feedback maps (the so-called lookup tables) and modifying the actuators.

Mechanically fixed and dynamically regulated by mechanical and pneumatic means, air-fuel combination, ignition time, and idle speed were before ECUs.

As soon as the system gets battery voltage, after ignition is turned, the efi computer makes a self-test of all the actuators and sensors, included the ETC.

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I need help with this pls
enyata [817]
Friction makes a moving object slower
8 0
3 years ago
a body accelerates uniformly from rest at 2m/s^2 for 5 seconds. Calculate its averege velocity in this time​
Lana71 [14]

HERE IS YOUR ANSWER!

8 0
3 years ago
Arrow_forward
garri49 [273]

Explanation:

(a) Hooke's law:

F = kx

7.50 N = k (0.0300 m)

k = 250 N/m

(b) Angular frequency:

ω = √(k/m)

ω = √((250 N/m) / (0.500 kg))

ω = 22.4 rad/s

Frequency:

f = ω / (2π)

f = 3.56 cycles/s

Period:

T = 1/f

T = 0.281 s

(c) EE = ½ kx²

EE = ½ (250 N/m) (0.0500 m)²

EE = 0.313 J

(d) A = 0.0500 m

(e) vmax = Aω

vmax = (0.0500 m) (22.4 rad/s)

vmax = 1.12 m/s

amax = Aω²

amax = (0.0500 m) (22.4 rad/s)²

amax = 25.0 m/s²

(f) x = A cos(ωt)

x = (0.0500 m) cos(22.4 rad/s × 0.500 s)

x = 0.00919 m

(g) v = dx/dt = -Aω sin(ωt)

v = -(0.0500 m) (22.4 rad/s) sin(22.4 rad/s × 0.500 s)

v = -1.10 m/s

a = dv/dt = -Aω² cos(ωt)

a = -(0.0500 m) (22.4 rad/s)² cos(22.4 rad/s × 0.500 s)

a = -4.59 m/s²

3 0
3 years ago
PLEASE HELP ME on this question.
VladimirAG [237]
The second one is the answer
4 0
4 years ago
What is the net work done on the 20kg block while it moves the 4 meters?
Vinil7 [7]

Answer:

The answer to your question is 784.8 J. None of your answer, did you forget some information?

Explanation:

Data

mass = 20 kg

distance = 4 m

work = ?

Formula

Work = force x distance

Force = mass x gravity

Process

1.- Calculate the weight of the block

     Weight = 20 x 9.81

     Weight = 196.2 N

2.- Calculate the work done

     Work = 196.2 x 4

     Work = 784.8 J

5 0
4 years ago
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