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3 years ago

What are some common ways of converting fuel to useful energy? Check all that apply.

Engineering

2 answers:

Pavlova-9 [17]3 years ago

5 0

Answer:

<h2>Burning it and heating it.</h2>

Explanation:

just saw it on edge.

Rainbow [258]3 years ago

3 0

Answer: Burning it.

Explanation: When your car moves its due to the burning of fuel, works like a turbine the molecules evaporate and turn a turbine like object.

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What is a shearing stress? Is there a force resulting from two solids in contact to which is it similar?

Luba_88 [7]

Answer:

Shearing stresses are the stresses generated in any material when a force acts in such a way that it tends to tear off the material.

Generally the above definition is valid at an armature level, in more technical terms shearing stresses are the component of the stresses that act parallel to any plane in a material that is under stress. Shearing stresses are present in a body even if normal forces act on it along the centroidal axis.

Mathematically in a plane AB the shearing stresses are given by

$\tau =\frac{Fcos(\theta )}{A}$

Yes the shearing force which generates the shearing stresses is similar to frictional force that acts between the 2 surfaces in contact with each other.

7 0

3 years ago

To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105

GenaCL600 [577]

Answer:

$62.14\ \text{miles}$

$6213727.37\ \text{miles}$

Explanation:

The distance of the chain would be the product of the dislocation density and the volume of the metal.

Dislocation density = $10^5\ \text{mm}^{-2}$

Volume of the metal = $1000\ \text{mm}^3$

$10^5\times 1000=10^8\ \text{mm}\\ =10^5\ \text{m}$

$1\ \text{mile}=1609.34\ \text{m}$

$\dfrac{10^5}{1609.34}=62.14\ \text{miles}$

The chain would extend $62.14\ \text{miles}$

Dislocation density = $10^{10}\ \text{mm}^{-2}$

Volume of the metal = $1000\ \text{mm}^3$

$10^{10}\times 1000=10^{13}\ \text{mm}\\ =10^{10}\ \text{m}$

$\dfrac{10^{10}}{1609.34}=6213727.37\ \text{miles}$

The chain would extend $6213727.37\ \text{miles}$

3 0

3 years ago

Air enters a compressor steadily at the ambient conditions of 100 kPa and 22°C and leaves at 800 kPa. Heat is lost from the comp

telo118 [61]

Answer:

a) 358.8K

b) 181.1 kJ/kg.K

c) 0.0068 kJ/kg.K

Explanation:

Given:

P1 = 100kPa

P2= 800kPa

T1 = 22°C = 22+273 = 295K

q_out = 120 kJ/kg

∆S_air = 0.40 kJ/kg.k

T2 =??

a) Using the formula for change in entropy of air, we have:

∆S_air = $c_p In \frac{T_2}{T_1} - Rln \frac{P_2}{P_1}$

Let's take gas constant, Cp= 1.005 kJ/kg.K and R = 0.287 kJ/kg.K

Solving, we have:

[/tex] -0.40= (1.005)ln\frac{T_2}{295} ln\frac{800}{100}[/tex]

$-0.40= 1.005(ln T_2 - 5.68697)- 0.5968$

Solving for T2 we have:

$T_2 = 5.8828$

Taking the exponential on the equation (both sides), we have:

[/tex] T_2 = e^5^.^8^8^2^8 = 358.8K[/tex]

b) Work input to compressor:

$w_in = c_p(T_2 - T_1)+q_out$

$w_in = 1.005(358.8 - 295)+120$

= 184.1 kJ/kg

c) Entropy genered during this process, we use the expression;

Egen = ∆Eair + ∆Es

Where; Egen = generated entropy

∆Eair = Entropy change of air in compressor

∆Es = Entropy change in surrounding.

We need to first find ∆Es, since it is unknown.

Therefore ∆Es = $\frac{q_out}{T_1}$

$\frac{120kJ/kg.k}{295K}$

∆Es = 0.4068kJ/kg.k

Hence, entropy generated, Egen will be calculated as:

= -0.40 kJ/kg.K + 0.40608kJ/kg.K

= 0.0068kJ/kg.k

3 0

3 years ago

Machine movement can be divided into what two main categories?

pishuonlain [190]

Answer:

motion and power

Explanation:

8 0

3 years ago

Read 2 more answers

Define the coefficient of determination and discuss the impact you would expect it to have on your engineering decision-making b

scoundrel [369]

Answer and Explanation:

The coefficient of determination also called "goodness of fit" or R-squared(R²) is used in statistical measurements to understand the relationship between two variables such that changes in one variable affects the other. The level of relationship or the degree to which one affects the other is measured by 0 to 1 whereby 0 means no relationship at all and 1 means one totally affects the other while figures in between such 0.40 would mean one variable affects 40% of the other variable.

In making a decision as an engineer while using the coefficient of determination, one would try to understand the relationship between variables under consideration and make decisions based on figures obtained from calculating coefficient of determination. In other words when there is a 0 coefficient then there is no relationship between variables and an engineer would make his decisions with this in mind and vice versa.

7 0

3 years ago