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3 years ago

Alcohol consumption tends to cause more ___________ behavior.

Engineering

1 answer:

lions [1.4K]3 years ago

8 0

Answer:

Aggressive behavior

Explanation:

Alcohol consumption tends to cause more Aggressive behavior.

The consumption of alcohol plays a more role in our culture but drinking of too much alcohol can cause drowsiness, vomiting, Upset stomach, slurred speech, heart damage, infertile, numbness lung infections, and many more. Also too much alcohol can cause violence, anger and so on in the society.

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Solar energy stored in large bodies of water, called solar ponds, is being used to generate electricity. If such a solar power p

fgiga [73]

Answer: 1.137*10^7 Btu/h.

Explanation:

Given data:

Efficiency of the plant = 4.5percent

Net power output of the plant = 150kw

Solution:

The required collection rate

QH = W/n

= 150/0.045 * 0.94782/ 1 /60 */60 Btu/h.

= 3333.333 *3412.152Btu/h.

= 11373840 Btu/h

= 1.137*10^7 Btu/h.

3 0

2 years ago

A bridge to be fabricated of steel girders is designed to be 500 m long and 12 m wide at ambient temperature (assumed 20°C). Exp

Volgvan

Answer:

a) 22.5number

b) 22.22 m length

Explanation:

Given data:

Bridge length = 500 m

width of bridge = 12 m

Maximum temperature = 40 degree C

minimum temperature = - 35 degree C

Maximum expansion can be determined as

$\Delta L = L \alpha (T_{max} - T_{min})$

where , \alpha is expansion coefficient $= 12\times 10^{-6}$ degree C

SO, $\Delta L = 500\times 12\times 10^{-6}\times ( 40 - (-35))$

$\Delta L = 0.45 m = 450 mm$

number of minimum expansion joints is calculated as

$n = \frac{450}{20} = 22.5$

b) length of each bridge

$Length = \frac{500}{22.5} = 22.22 m$

8 0

2 years ago

Which of the following is not a function of the cooling system

Pie

Answer:

hola soy dora

Explanation:

5 0

2 years ago

The dam cross section is an equilateral triangle, with a side length, L, of 50 m. Its width into the paper, b, is 100 m. The dam

lisabon 2012 [21]

Answer:

Explanation:

In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.

We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.

The weigth depends on the size and specific gravity.

W = 1/2 * b * h * L * SG

Then

Teq = 1/2 * b * h * L * SG * b / 2

Teq = 1/4 * b^2 * h * L * SG

The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:

$T1 = \int\limits^h_0 {p(y) * sin(30) * L * (h-y)} \, dy$

The term sin(30) is because of the slope of the wall

The pressure of water is:

p(y) = SGw * (h - y)

Then:

$T1 = \int\limits^h_0 {SGw * (h-y) * sin(30) * L * (h-y)} \, dy$

$T1 = \int\limits^h_0 {SGw * sin(30) * L * (h-y)^2} \, dy$

$T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy$

$T1 = SGw * sin(30) * L * \int\limits^h_0 {h^2 - 2*h*y + y^2} \, dy$

T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)

T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)

T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)

T1 = 1/3 * SGw * sin(30) * L * h^3

To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)

1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3

In an equilateral triangle h = b * cos(30)

1/4 * b^3 * cos(30) * L * SG > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3

SG > SGw * 4/3* sin(30) * (cos(30))^2

SG > 1/2 * SGw

For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.

This is avergae specific gravity, including holes.

6 0

2 years ago

Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essentially t

zepelin [54]

Answer:

attached below

Explanation:

4 0

3 years ago