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3 years ago

What regulates the car engines temperature

Engineering

2 answers:

velikii [3]3 years ago

8 0

Answer:

car’s thermostat is used to regulate the temperature of the engine to help the engine stay cool.

Explanation:

vlada-n [284]3 years ago

3 0

Answer:

Coolant

Explanation:

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Peter the postman became bored one night and, to break the monotony of the night shift, he carried out the following experiment

aleksandrvk [35]

Answer:

//This Program is written in C++

// Comments are used for explanatory purpose

#include <iostream>

using namespace std;

enum mailbox{open, close};

int box[149];

void closeAllBoxes();

void OpenClose();

void printAll();

int main()

{

closeAllBoxes();

OpenClose();

printAll();

return 0;

}

void closeAllBoxes()

{

for (int i = 0; i < 150; i++) //Iterate through from 0 to 149 which literarily means 1 to 150

{

box[i] = close; //Close all boxes

}

void OpenClose()

{

for(int i = 2; i < 150; i++) {

for(int j = i; j < 150; j += i) {

if (box[j] == close) //Open box if box is closed

box[j] = open;

else

box[j] = close; // Close box if box is opened

}

// At the end of this test, all boxes would be closed

}

void printAll()

{

for (int x = 0; x < 150; x++) //use this to test

{

if (box[x] = 1)

{

cout << "Mailbox #" << x+1 << " is closed" << endl;

// Print all close boxes

}

Explanation:

7 0

3 years ago

A two-dimensional flow field described by

Oduvanchick [21]

Answer:

the answer is

Explanation:

<h2> We now focus on purely two-dimensional flows, in which the velocity takes the form </h2><h2>u(x, y, t) = u(x, y, t)i + v(x, y, t)j. (2.1) </h2><h2>With the velocity given by (2.1), the vorticity takes the form </h2><h2>ω = ∇ × u = </h2><h2> </h2><h2>∂v </h2><h2>∂x − </h2><h2>∂u </h2><h2>∂y </h2><h2>k. (2.2) </h2><h2>We assume throughout that the flow is irrotational, i.e. that ∇ × u ≡ 0 and hence </h2><h2>∂v </h2><h2>∂x − </h2><h2>∂u </h2><h2>∂y = 0. (2.3) </h2><h2>We have already shown in Section 1 that this condition implies the existence of a velocity </h2><h2>potential φ such that u ≡ ∇φ, that is </h2><h2>u = </h2><h2>∂φ </h2><h2>∂x, v = </h2><h2>∂φ </h2><h2>∂y . (2.4) </h2><h2>We also recall the definition of φ as </h2><h2>φ(x, y, t) = φ0(t) + Z x </h2><h2>0 </h2><h2>u · dx = φ0(t) + Z x </h2><h2>0 </h2><h2>(u dx + v dy), (2.5) </h2><h2>where the scalar function φ0(t) is arbitrary, and the value of φ(x, y, t) is independent </h2><h2>of the integration path chosen to join the origin 0 to the point x = (x, y). This fact is </h2><h2>even easier to establish when we restrict our attention to two dimensions. If we consider </h2><h2>two alternative paths, whose union forms a simple closed contour C in the (x, y)-plane, </h2><h2>Green’s Theorem implies that </h2><h2> </h2><h2> </h2><h2> </h2><h2> </h2><h2> </h2><h2></h2><h2></h2>

5 0

3 years ago

Precast double-tee units are generally used as structural floors when:________.

TiliK225 [7]

Answer:

<em>d. floor spans exceed that which is economical with site-cast or hollow-core slabs and exceed 60 feet.</em>

Explanation:

<em> Double tee slabs are prestressed concrete castings usually about 100 m to 150 m long</em>. <em>They are usually used in the construction industry in structures requiring a long floor span (or roof span) exceeding 60 feet in length, that is required to bear additional load (additional load carrying capability</em>). Double tee slabs are usually employed in construction because they shorten construction time. reduce congestion at the construction site, and also reduce construction cost.

<em>They are the required choice in construction when the cost of site casting or hollow-core slabs is no longer economical due to the length of the structure exceeding 60 feet.</em>

8 0

3 years ago

A four-lane divided multilane highway (two lanes in each direction) in rolling terrain has five access points per mile and 11-ft

grandymaker [24]

Answer:

peak-hour volume = 1890 veh/h

Explanation:

<u>Determine the peak-hour Volume </u>

Applying the equation below

Vp = v / ( PHF * N * Fg * Fdp ) -------------- ( 1 )

where :

Vp = 1250

v ( peak - hour volume ) = ?

PHF ( peak hour factor ) = 0.84

N = 2 lanes per direction

Fg ( grade adjustment for rolling terrain ) = 0.99 ≈ 1

Fdp = 0.90

<u>Back to equation 1 </u>

v = Vp ( PHF * N * Fg * Fdp )

= 1250 ( 0.84 * 2 * 1 * 0.90 )

= 1890 veh/h

5 0

3 years ago

What is the mass in both slugs and kilograms of a 1000-lb beam?

Likurg_2 [28]

Answer:

31.1 slug, 453.4 Kg

Explanation:

given,

mass of the beam is 1000 lb

to convert mass of beam into slugs and kilograms.

1 lb is equal to 0.0311 slug

1000 lb = 1000 × 0.0311

= 31.1 slug

now, conversion of lb into kg

1 lb is equal to 0.4534 kg

so,

1000 lb = 1000 × 0.4534

= 453.4 Kg

hence, 1000 lb of beam in slugs is equal to 31.1 slugs and in kilo gram is 453.4 Kg.

7 0

3 years ago