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3 years ago

Describe carbonation as it applies to the four-stroke engine.

Engineering

2 answers:

valentinak56 [21]3 years ago

5 0

Carbonation is more of a healer to the engine

erastovalidia [21]3 years ago

4 0

Carbon is a healer and it applies to many helpful things

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Find the total amount of heat in Q lost through a wall 10' by 18' , with R value from q. 1. Inside temperature is 70 degrees F w

marissa [1.9K]

Answer:

Just think

Explanation:

6 0

4 years ago

Read 2 more answers

A glass plate is subjected to a tensile stress of 40 MPa. If the specific surface energy is 0.3 J/m2 and the modulus of elastici

Alika [10]

Answer:

The maximum length of a surface flaw is 8.24 μm

8.24 μm

Explanation:

Given that:

The modulus of elasticity E = 69 GPa

The specific surface energy $\delta_s$ = 0.3 J/m²

The length of the surface flaw "a" = ??

From the theory of the brittle fracture;

$\sigma _c = \bigg ( \dfrac{2E \delta_s}{\pi a} \bigg )^{1/2}$

Making a the subject of the formula; we have:

$a = \bigg ( \dfrac{2 \times E \times \delta_s}{\pi \sigma _c ^2} \bigg )$

$a= \bigg ( \dfrac{2 \times 69*10^9 \times 0.3}{\pi (40*10^6)^2} \bigg )$

a = 8.24 × 10⁻⁶ m

a = 8.24 μm

Thus; the maximum length of a surface flaw is 8.24 μm

3 0

3 years ago

Which option identifies the type of power system Tommy will design in the following scenario?

Sedaia [141]

Answer:

diagram of an electrical curcuit

an sketch of an HVAC system

Also 3D image of a hydrualic piston

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3 years ago

Blocks A and B each have a mass m. Determine the largest horizontal force P which can be applied to B so that A will not move re

slava [35]

Answer:

The answer is "15 N".

Explanation:

Please find the complete question in the attached file.

In frame B:

For just slipping:

$\to \frac{P}{2} \cos \theta =mg \sin \theta\\\\\to P=2 mg \tan \theta \\\\$

$=2 \times 1 \times g \times \tan 37^{\circ}\\\\ =2 \times 10 \times \frac{3}{4}\\\\ =15 \ N$

4 0

3 years ago

Nancy ate a 500 Cal lunch. Neglecting efficiency issues (i.e., assuming 100% conversion of energy to work), to what height could

VladimirAG [237]

Answer:

$4265.04\ \text{m}$

$2.38\times 10^{10}\ \text{W}$

Explanation:

PE = Energy of food = 500 cal = $500\times4184=2.092\times10^6\ \text{J}$

m = Mass of object = 50 kg

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

Potential energy of food is given by

$PE=mgh\\\Rightarrow h=\dfrac{PE}{mg}\\\Rightarrow h=\dfrac{2.092\times 10^6}{50\times 9.81}\\\Rightarrow h=4265.04\ \text{m}$

Nancy could raise the weight to a maximum height of $4265.04\ \text{m}$ .

Mass of $H_2$ used per year = $25\times 10^{9}\ \text{kg/year}$

Energy of $H_2$ = $\dfrac{30\times10^9}{1000}=30\times 10^6\ \text{J/kg}$

Power

$P=25\times 10^{9}\ \text{kg/year}\times 30\times 10^6\ \text{J/kg}\\\Rightarrow P=7.5\times 10^{17}\ \text{J/year}\\\Rightarrow P=\dfrac{7.5\times 10^{17}}{365.25\times 24\times 60\times 60}\\\Rightarrow P=2.38\times 10^{10}\ \text{W}$

The power requirement is $2.38\times 10^{10}\ \text{W}$ .

6 0

3 years ago