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3 years ago

Which option demonstrates when most vehicles lose their efficiency?

Engineering

2 answers:

faltersainse [42]3 years ago

7 0

Answer:

During starts and stops

Explanation:

"Many of the energy inefficiencies in vehicles occur during starts and stops."

found it in the text.

guapka [62]3 years ago

6 0

Hey there ..

I think the answer is as they put brake ..

I am not sure .. just check the image also provided above .. ..

If u think this helped u ..plz mark me as brainliest ..

And follow me

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The minimum recommended standards for the operating system, processor, primary memory (RAM), and storage capacity for certain so

nlexa [21]

Answer:System requirements

Explanation:

its right.

8 0

3 years ago

Read 2 more answers

A saturated 1.5 ft3 clay sample has a natural water content of 25%, shrinkage limit (SL) of 12% and a specific gravity (GS) of 2

Svetllana [295]

$79 f t^{3}$ is the volume of the sample when the water content is 10%.

<u>Explanation:</u>

Given Data:

$V_{1}=100\ \mathrm{ft}^{3}$

First has a natural water content of 25% = $\frac{25}{100}$ = 0.25

Shrinkage limit, $w_{1}=12 \%=\frac{12}{100}=0.12$

$G_{s}=2.70$

We need to determine the volume of the sample when the water content is 10% (0.10). As we know,

$V \propto[1+e]$

$\frac{V_{2}}{V_{1}}=\frac{1+e_{2}}{1+e_{1}}$ ------> eq 1

$e_{1}=\frac{w_{1} \times G_{s}}{S_{r}}$

The above equation is at $S_{r}=1$ ,

$e_{1}=w_{1} \times G_{s}$

Applying the given values, we get

$e_{1}=0.25 \times 2.70=0.675$

Shrinkage limit is lowest water content

$e_{2}=w_{2} \times G_{s}$

Applying the given values, we get

$e_{2}=0.12 \times 2.70=0.324$

Applying the found values in eq 1, we get

$\frac{V_{2}}{100}=\frac{1+0.324}{1+0.675}=\frac{1.324}{1.675}=0.7904$

$V_{2}=0.7904 \times 100=79\ \mathrm{ft}^{3}$

7 0

3 years ago

Complete function PrintPopcornTime(), with int parameter bagOunces, and void return type. If bagOunces is less than 3, print "To

weqwewe [10]

Answer:

#include <iostream>

using namespace std;

void PrintPopcornTime(int bagOunces) {

if(bagOunces < 3){

cout << "Too small";

cout << endl;

}

else if(bagOunces > 10){

cout << "Too large";

cout << endl;

}

else{

cout << (6 * bagOunces) << " seconds" << endl;

}

int main() {

PrintPopcornTime(7);

return 0;

}

Explanation:

Using C++ to write the program. In line 1 we define the header "#include <iostream>" that defines the standard input/output stream objects. In line 2 "using namespace std" gives me the ability to use classes or functions, From lines 5 to 17 we define the function "PrintPopcornTime(), with int parameter bagOunces" Line 19 we can then call the function using 7 as the argument "PrintPopcornTime(7);" to get the expected output.

8 0

3 years ago

10. An engineer is designing a total hip implant. She intends to make the femoral stem out of titanium because it forms a good i

creativ13 [48]

Answer:

Yes. She should be worried about corrosion. The 18-8 stainless exhibits intergranular corrosion due to high (0.08%) carbon content and gross pitting due to low molybdenum content.

Explanation: lol

8 0

4 years ago

An adiabatic air compressor compresses 10 L/s of air at 120 kPa and 20 degree C to 1000 kPa and 300 degree C.

Oksana_A [137]

Answer:

work=281.4KJ/kg

Power=4Kw

Explanation:

Hi!

To solve follow the steps below!

1. Find the density of the air at the entrance using the equation for ideal gases

$density=\frac{P}{RT}$

where

P=pressure=120kPa

T=20C=293k

R= 0.287 kJ/(kg*K)= gas constant ideal for air

$density=\frac{120}{(0.287)(293)}=1.43kg/m^3$

2.find the mass flow by finding the product between the flow rate and the density

m=(density)(flow rate)

flow rate=10L/s=0.01m^3/s

m=(1.43kg/m^3)(0.01m^3/s)=0.0143kg/s

3. Please use the equation the first law of thermodynamics that states that the energy that enters is the same as the one that must come out, we infer the following equation, note = remember that power is the product of work and mass flow

Work

w=Cp(T1-T2)

Where

Cp= specific heat for air=1.005KJ/kgK

w=work

T1=inlet temperature=20C

T2=outlet temperature=300C

w=1.005(300-20)=281.4KJ/kg

Power

W=mw

W=(0.0143)(281.4KJ/kg)=4Kw

5 0

3 years ago