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3 years ago

Thoughts about drinking and driving

Engineering

2 answers:

poizon [28]3 years ago

8 0

Answer: it’s very bad behavior and can be dangerous

Explanation:

Marysya12 [62]3 years ago

8 0

Answer:

don't drink while you drive

Explanation:

don't drink while you drive because when your drinking while u drive you can crash and be in a big accident.

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Technician A says that the distributor cap provides a connection point between the rotor and each individual cylinder plug wire.

Vera_Pavlovna [14]

Answer:

Neither a or b.

Explanation:

The distributor cap is cover which protects internal parts and holds contact between internal rotor and spark plug wires. When inspecting distributor caps there should be some carbon tracking which looks like bright white grease trails.

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3 years ago

An agricultural manager requires

sp2606 [1]

C, being able to maintain legal information on grant programs

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3 years ago

Design a posttest-only experiment that would test each of the following causal claims. For each one, identify the study’s indepe

Lostsunrise [7]

Answer:

Here independent variable is negative reinforcement like punishment. The dependent variable is aggressiveness.

Control group : Teacher

Experiment group : Students

Independent variable : Exercise

Dependent variable : Depression

Various levels of independent variables : The time of exercise may be increased or reduced

Independent variable: Environment friendly messages

Dependent variables : Environment friendly behavior

Interaction: The verbal exchange among the peers.

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3 years ago

A thick spherical pressure vessel of inner radius 150 mm is subjected to maximum an internal pressure of 80 MPa. Calculate its w

ivann1987 [24]

Answer:

by principal stress theory

t = 20.226

by total strain theory

t = 20.36

Explanation:

given data

internal radius $r_{1}$ = 150 mm

pressure p = 80 MPa

yield strength = 300 MPa

poisson's ratio = 0.3

a) by principal stress theory

thickness can be obtained as t

t = $r_{1}\left [ (\frac{\sigma _{y} +p}{\sigma _{y} - 0.5p})^{1/3}-1 \right ]$

t = = 150\left [ (\frac{300 +80}{300-0.5*80})^{1/3}-1 \right ]

t = 20.226

b) by total strain theory

m = $\frac{\sigma _{y}}{p}$

m = $\frac{300}{80}$ = 3.75

we know that

K = $\frac{r_{2}}{r_{1} }$

$\frac{K^{3}+1}{K^{3}-1}= \frac{-2\mu +\sqrt{4\mu^{2}-2(1-\mu)(1-m^{^{2}}))}}{1-\mu}$

$\frac{K^{3}+1}{K^{3}-1}= \frac{-2*0.3 +\sqrt{4*0.3^{2}-2(1-0.3)(1-3.75^{^{2}}))}}{1-0.3}$

$\frac{K^{3}+1}{K^{3}-1}= 5.3$

k = 1.13

1.13 = $\frac{r_{2}}{150 }$

$r_{2}$ = 170.36 mm

t = $r_{2}$ - $r_{1}$

t = 170.36 - 150

t = 20.36

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4 years ago

A 10-ft-long simply supported laminated wood beam consists of eight 1.5-in. by 6-in. planks glued together to form a section 6 i

ruslelena [56]

Answer:

point B where $Q_B = 101.25 \ in^3$ has the largest Q value at section a–a

Explanation:

The missing diagram that is suppose to be attached to this question can be found in the attached file below.

So from the given information ;we are to determine the point that has the largest Q value at section a–a

In order to do that; we will work hand in hand with the image attached below.

From the image attached ; we will realize that there are 8 blocks aligned on top on another in the R.H.S of the image with the total of 12 in; meaning that each block contains 1.5 in each.

We also have block partitioned into different point segments . i,e A,B,C, D

For point A ;

Let Q be the moment of the Area A;

SO ; $Q_A = Area \times y_1$