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mezya [45]
3 years ago
11

Sarah and Raj take/takes me to a baseball game every year.

Engineering
1 answer:
timofeeve [1]3 years ago
4 0

Answer:

take

play

are

sits

crouches

direct

indirect

direct

direct

indirect

Explanation:

i had just taken the test

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DO you of you guys know about Any Anmial rights it is for a resercah project
gizmo_the_mogwai [7]

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.

4 0
3 years ago
Find the following for an input of 120 VAC(RMS), 60 hertz, given a 10:1 stepdown transformer, and a full-wave bridge rectifier.
atroni [7]

Answer:

(i) 169.68 volt

(ii) 16.90 volt

(iii) 16.90 volt

(iv) 108.07 volt

(v) 2.161 A

Explanation:

Turn ratio is given as 10:1

We have given that input voltage v_p=120volt

(i) We know that peak voltage is give by v_{peak}=\sqrt{2}v_p=\sqrt{2}\times 120=169.68volt

(ii) We know that for transformer \frac{v_p}{v_s}=\frac{n_p}{n_s}

So \frac{169.08}{v_s}=\frac{10}{1}

v_s=16.90volt

So peak voltage in secondary will be 16.90 volt

(iii) Peak voltage of the rectifier will be equal to the peak voltage of the secondary

So peak voltage of the rectifier will be 16.90 volt

(iv) Dc voltage of the rectifier is given by v_{dc}=\frac{2v_m}{\pi }=\frac{2\times 1.414\times 120}{3.14}=108.07volt

(v) Now dc current is given by i_{dc}=\frac{v_{dc}}{R}=\frac{108.07}{50}=2.1614A

4 0
3 years ago
What is made in heaven?​
kramer

Answer:

Babies come from heaven didn't you know?

3 0
2 years ago
Explain combined normal and shear stresses with sketch. Write the general expression for (a) Normal and shear stresses on inclin
Alborosie

Answer:

a) Normal stress :

бn =[ ( бx + бy ) / 2  + ( бx - бy ) / 2  ] cos2∅ + Txysin2∅

shear stress

Tn = ( - бx - бy ) / 2  sin2∅ + Txy cos2∅

b) principal stress :

б1 = ( бx + бy ) / 2  - \sqrt{}( ( бx - бy ) / 2 )^2 + T^2xy

maximum shear stress:

Tmax  = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy

Explanation:

Combined normal stress and shear stress  sketches attached below

The terms in the sketch are :

бx = tensile stress in x direction

бy =  tensile stress in y direction

Txy = y component of shear stress acting on the perpendicular plane to x axis

бn = Normal stress acting on the inclined plane EF

Tn = shear stress acting on the inclined plane EF

A) Normal and shear stresses on inclined plane

Normal stress :

бn =[ ( бx + бy ) / 2  + ( бx - бy ) / 2  ] cos2∅ + Txysin2∅

shear stress

Tn = ( - бx - бy ) / 2  sin2∅ + Txy cos2∅

B) principal and maximum shear stresses

principal stress :

б1 = ( бx + бy ) / 2  - \sqrt{}( ( бx - бy ) / 2 )^2 + T^2xy

maximum shear stress:

Tmax  = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy

6 0
2 years ago
Give the approximate temperature (in K) at which creep deformation becomes an important consideration for each of the following
andrezito [222]

Answer:

691K, 543K, 725K, 1473K, 240K, 373K

Explanation:

Creep deformation of any metal is the transformational tendency of a metal to distort rapidly or slowly when attacked by any form of mechanical stress. The temperature significant for a metal to deform is gotten by the division of the actual temperature of the metal by its melting point. This is termed homologous temperature which is 0.4 or higher. It is calculated by the equation:

0.4Tm

Therefore for the listed metals...

For Nickel, 0.4Tm = 0.4 ×(1455 + 273) = 691 K

For Copper, 0.4Tm = 0.4 ×(1085 + 273) = 543 K

For Iron, 0.4Tm = 0.4 ×(1538 + 273) = 725 K

For Tungsten, 0.4Tm = 0.4 ×(3410 + 273) = 1473 K

For Lead, 0.4Tm = 0.4 × (327 + 273) = 240 K

For Aluminium, 0.4Tm = 0.4 ×(660 + 273) = 373 K

5 0
3 years ago
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