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3 years ago

DO you of you guys know about Any Anmial rights it is for a resercah project

Engineering

1 answer:

gizmo_the_mogwai [7]3 years ago

4 0

Explanation:

Sure! I'll give you the definition to start off-

The right of animals to be free from exploitation, domination and abuse by humans. Free-living Animals & Their Environment; To live free, animals need a place to live. Wildlife Law Program; The Wildlife Law Program focuses on the defense of wildlife and their habitats throughout the world.

Here are some examples of companies for Animal Advocates for inspiration.

PETA – People for the Ethical Treatment of Animals

International Fund for Animal Welfare – IFAW

Cincinnati Zoo & Botanical Garden

Here are some of the basic rights of animals.

- No experiments on animals.

- No breeding and killing animals for food or clothes or medicine.

- No use of animals for hard labor.

- No selective breeding for any reason other than the benefit of the animal.

- No hunting.

- No zoos or use of animals in entertainment.

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What are the four basic parts of process plan

Rus_ich [418]

Answer:

1. Goal setting

2. Developing the planning premises

3. Reviewing Limitations

4. Deciding the planning period

5. Formulation of policies and strategies

6. Preparing operating plans

7. Integration of plans

Explanation:

I don't have an explanation but i thru in a few extra answers for like extra credit or something if your teacher does that

8 0

3 years ago

A vehicle of 1 200 kg is moving at a speed of 40 km/h on an incline of 1 in 50. The total constant rolling and wind resistance i

Readme [11.4K]

Answer:11.602 KW

Explanation:

mass of vehicle $\left ( m\right )=1200 kg$

speed=40Km/h

Resistance=600 N

$\eta =80%$

Gear ratio $\left ( G\right )=4:1$

$D_{effective}=500mm$

Net force to overcome by engine is

F=Resistance + sin component of weight

F=600+ $mgsin\theta$

Where $tan\theta =[tex]\frac{1}{50}$

$\theta =1.1457^{\circ}$

$F=600+1200\times 9.81\times sin\left ( 1.1457\right )$

F=600+235.38=835.38 N

power= $F.v=835.38\frac{100}{9}$

Engine Power= $\frac{835.\frac{100}{9}}{\eta }$ =11.602 KW

4 0

4 years ago

(Laminar flow) A fluid flows through two horizontal pipes of equal length which are connected together to form a pipe of length

EastWind [94]

This question is incomplete, the complete question is;

(Laminar flow) A fluid flows through two horizontal pipes of equal length which are connected together to form a pipe of length 2 $l$ . The flow is laminar and fully developed. The pressure drop for the first pipe is 1.657 times greater than it is for the second pipe. If the diameter of the first pipe is D, determine the diameter of the second pipe.

D₃ = _____D.

{ the tolerance is +/-3% }

Answer:

the diameter of the second pipe D₃ is 1.13D

Explanation:

Given the data in the question;

Length = 2 $l$

pressure drop in the first pipe is 1.657 times greater than it is for the second pipe.

Now, we know that for Laminar Flow;

V' = πD⁴ΔP / 128μL

where V'₁ = V'₂ and ΔP₁₋₂ = 1.657 ΔP₂₋₃

Hence,

V'₁ = πD⁴ΔP₁₋₂ / 128μL = V'₃ = πD₃⁴ΔP₂₋₃ / 128μL

D₃ = D $($ ΔP₁₋₂ / ΔP₂₋₃ $)^{\frac{1}{4}$

we substitute

D₃ = D $($ 1.657 $)^{\frac{1}{4}$

D₃ = D( 1.134568 )

D₃ = 1.13D

Therefore, the diameter of the second pipe D₃ is 1.13D

8 0

3 years ago

A laboratory in the Y building keep a vacuum pressure of 0.1 kPa abs. What is the net force acting on the door considering the a

seropon [69]

Answer:

net force acting on the floor is 100 kN

Explanation:

Given data:

$P_{vaccum} = 0.1 kPa$

$P_{atm} = 101.325 kPa$

dimension of floor = 2 m \times 0.5 m

we know that

Net force can be calculated as follow

$f_{net} = P_{vaccum} \times area$

$f_{net} = 0.1\times 10^3 \times 2\times 0.5$

$f_{net} = 0.1\times 10^3 \times 1$

$f_{net} = 100 kN$

Therefore net force acting on the floor is 100 kN

7 0

3 years ago

A cylinder with a 6.0 in. diameter and 12.0 in. length is put under a compres-sive load of 150 kips. The modulus of elasticity f

jeka94

Answer:

Final Length = 11.992 in

Final Diameter = 6.001 in

Explanation:

First we calculate the cross-sectional area:

Area = A = πr² = π(3 in)² = 28.3 in²

Now, we calculate the stress:

Stress = Compressive Load/Area

Stress = - 150 kips/28.3 in²

Stress = -5.3 ksi

Now,

Modulus of Elasticity = Stress/Longitudinal Strain

8000 ksi = -5.3 ksi/Longitudinal Strain

Longitudinal Strain = -6.63 x 10⁻⁴

but,

Longitudinal Strain = (Final Length - Initial Length)/Initial Length

-6.63 x 10⁻⁴ = (Final Length - 12 in)/12 in

Final Length = (-6.63 x 10⁻⁴)(12 in) + 12 in

<u>Final Length = 11.992 in</u>

we know that:

Poisson's Ratio = - Lateral Strain/Longitudinal Strain

0.35 = - Lateral Strain/(- 6.63 x 10⁻⁴)

Lateral Strain = (0.35)(6.63 x 10⁻⁴)

Lateral Strain = 2.32 x 10⁻⁴

but,

Lateral Strain = (Final Diameter - Initial Diameter)/Initial Diameter

2.32 x 10⁻⁴ = (Final Diameter - 6 in)/6 in

Final Diameter = (2.32 x 10⁻⁴)(6 in) + 6 in

<u>Final Diameter = 6.001 in</u>

8 0

3 years ago