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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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Even when the head is held erect, as in the figure below, its center of mass is not directly over the principal point of support
alexandr1967 [171]

We are asked to determine the force required by the neck muscle in order to keep the head in equilibrium. To do that we will add the torques produced by the muscle force and the weight of the head. We will use torque in the clockwise direction to be negative, therefore, we have:

\Sigma T=r_{M\perp}(F_M)-r_{W\perp}(W)

Since we want to determine the forces when the system is at equilibrium this means that the total sum of torque is zero:

r_{M\perp}(F_M)-r_{W\perp}(W)=0

Now, we solve for the force of the muscle. First, we add the torque of the weight to both sides:

r_{M\perp}(F_M)=r_{W\perp}(W)

Now, we divide by the distance of the muscle:

(F_M)=\frac{r_{W\perp}(W)}{r_{M\perp}}

Now, we substitute the values:

F_M=\frac{(2.4cm)(50N)}{5.1cm}

Now, we solve the operations:

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Part B. To determine the force on the pivot we will add the forces we add the vertical forces:

\Sigma F_v=F_j-F_M-W

Since there is no vertical movement the sum of vertical forces is zero:

F_j-F_M-W=0

Now, we add the force of the muscle and the weight to both sides to solve for the force on the pivot:

F_j=F_M+W

Now, we plug in the values:

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F_j=73.53N

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8 0
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Answer:

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Answer:

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When padding is there, the time interval of contact is large and thus, the force exerted by the body is small.

So, when a person falls on the tile floor, there is no compression and thus, the time of contact is very small and thus the impulsive force is very large, due to  which the body may damage.

So, when a person falls on the carpeted floor, there is a compression and thus, the time of contact is comparatively large and thus the impulsive force is small, due to  which the body may safe.

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