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aivan3 [116]
2 years ago
12

You need to lift a 2012 toyota highlander and haven't done so in a while. which of these are reliable sources for checking the c

orrect lifting points?
Engineering
1 answer:
grandymaker [24]2 years ago
3 0

The reliable sources for checking the correct lifting points are: one need to check under the car's rocker panels, that is behind the front wheels and also a little bit ahead of the rear wheels.

<h3>How do you know where the proper lift points for a vehicle are?</h3>

The jacking points for a lot of vehicles are known to be reinforced metal ribs that are said to be designed so that it can be able to safely lift the vehicle.

Note that On a lot of vehicles, there are four jacking points. They are known to be seen seen at the located point that can be found under the car's rocker panels.

Hence, The reliable sources for checking the correct lifting points are: one need to check under the car's rocker panels, that is behind the front wheels and also a little bit ahead of the rear wheels.

Learn more about lifting points from

brainly.com/question/26209572

#SPJ1

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A motorist enters a freeway at 25 mi/h and accelerates uniformly to 65 mi/h. From the odometer in the car, the motorist knows th
Helga [31]

Answer:

a) 2.2 m/s² b) 8 s

Explanation:

a) Assuming that the acceleration is constant, we can use any of the kinematic equations to solve the question.

As we don´t know the time needed to accelerate, we can use the following equation:

vf2 – vo2 = 2*a*∆x

At first, we can convert the values of vf, vo and ∆x, to SI units, as follows:

vf = 65 mi/h* (1,605 m / 1mi) * (1h/3,600 sec) = 29 m/s

vo = 25 mi/h *(1,605 m / 1mi) * (1h/3,600 sec) = 11.2 m/s

∆x = 0.1 mi*(1,605 m / 1mi) = 160.5 m

Replacing these values in (1), and solving for a, we have:

a = (29 m/s – 11.2 m/s) / 321 m = 2.2 m/s2

b) In order to obtain the time needed to reach to 65 mi/h, we can rearrange the equation for the definition of acceleration, as follows:

vf = vo + at  

Replacing by the values already known for vo, vf and a, and solving for t, we get:

t = vf-vo /a = (29 m/s – 11.2 m/s) / 2.2 m/s = 8 sec

5 0
3 years ago
Refers to the capability to keep moving forward on a specified grade.
Mademuasel [1]

Answer:

maneuverability

Explanation:

needless to say, I took the quiz

6 0
3 years ago
Recommend the types of engineers needed to collaborate on a city project to build a skateboard park near protected wetlands.
4vir4ik [10]

Answer:

A civil engineer.

Explanation:

Civil engineering is the science that deals with the design, creation and maintenance of constructions for civil use on the earth's surface. Thus, this specialty seeks to adapt soils to the needs of life in society, creating buildings, bridges, and all other constructions adapted to civil life, while taking care of the correct use of soils and the correct distribution of spaces and resources. to be used for such constructions.

4 0
3 years ago
Say you have a random, unordered list containing 4096 four-digit numbers. Describe the most efficient way to: sort the list and
Debora [2.8K]

Answer:

Answer explained below

Explanation:

It is given that numbers are four-digit so maximum value of a number in this list could be 9999.

So we need to sort a list of integers, where each integer lies between [0,9999].

For these given constraints we can use counting sort which will run in linear time i.e. O(n).

--------------------------------------------------------------------------------

Psuedo Code:

countSort(int numList[]) {

int count[10000];

count[i] = 0; for all i;

for(int num in numList){

count[num]+= 1;

}

return count;

}

--------------------------------------------------------------------------------

Searching in this count array will be just O(1).

E.g. Lets say we want to search if 3 was present in the original list.

Case 1: it was present in the original list:

Then the count[3] would have been incremented by our sorting algorithm. so in case element exists then count value of that element will be greater than 0.

Case 2: it was not present:

In this case count[3] will remain at 0. so in case element does not exist then count of that element will be 0.

So to search for an element, say x, we just need to check if count[x]>0.

So search is O(1).

Run times:

Sorting: O(n)

Search: O(1)

6 0
3 years ago
Explain by Research how a basic generator works ? using diagram<br>​
natulia [17]
Correcto no se muy bien de que se trata el tema porque está en inglés.
Sorry
8 0
3 years ago
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