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2 years ago

A series of concrete pillars have been built on the border between Kuwait and Iraq. They are there to __________ the border.

1 answer:

5 0

A series of concrete pillars have been built on the border between Kuwait and Iraq. They are there to demarcate the border.

<h3>What is a border?</h3>

A border is a geographical boundary that separate<em> countries, states, provinces, counties, cities, and towns.</em>

A series of concrete pillars have been built on the border between Kuwait and Iraq. They are there to demarcate the border.

Find out more on border at: brainly.com/question/811755

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What would be the structure for the body points for a persuasive presentation?.

Nataliya [291]

A persuasive speech is structured like an informative speech. It has an introduction with an attention-getter and a clear thesis statement. It also has a body where the speaker presents their main points and it ends with a conclusion that sums up the main point of the speech.

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2 years ago

On highways, the far left lane is usually the _____. A. emergency lane B. merge lane C. slowest D. fastest

slamgirl [31]

On highways, the far left lane is usually the<u> fastest</u> moving traffic.

Answer: Option D.

<u>Explanation:</u>

For the most part, the right lane of a freeway is for entering and leaving the traffic stream. It is an arranging path, for use toward the start and end of your interstate run. The center paths are for through traffic, and the left path is for passing. On the off chance that you are not passing somebody, try not to be driving in the left path.

Regular practice and most law on United States expressways is that the left path is saved for passing and quicker moving traffic, and that traffic utilizing the left path must respect traffic wishing to surpass.

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3 years ago

Read 2 more answers

An aluminum alloy tube with an outside diameter of 3.50 in. and a wall thickness of 0.30 in. is used as a 14 ft long column. Ass

slega [8]

Answer:

slenderness ratio = 147.8

buckling load = 13.62 kips

Explanation:

Given data:

outside diameter is 3.50 inc

wall thickness 0.30 inc

length of column is 14 ft

E = 10,000 ksi

moment of inertia $= \frac{\pi}{64 (D_O^2 -D_i^2)}$

$I = \frac{\pi}{64}(3.5^2 -2.9^2) = 3.894 in^4$

Area $= \frac{\pi}{4} (3.5^2 -2.9^2) = 3.015 in^2$

$radius = \sqrt{\frac{I}{A}}$

$r = \sqrt{\frac{3.894}{3.015}$

r = 1.136 in

slenderness ratio $= \frac{L}{r}$

$= \frac{14 *12}{1.136} = 147.8$

buckling load $= P_cr = \frac{\pi^2 EI}}{l^2}$

$P_{cr} = \frac{\pi^2 *10,000*3.844}{( 14\times 12)^2}$

$P_{cr} = 13.62 kips$

3 0

2 years ago

soccer is also called association football" A soccer ball is a sphere, with circumference of 70 centimeters. in developing a new

timama [110]

Answer: Weight on Mars = 0.02593N

Explanation:

Given; Circumference C of Sphere = 70cm = 0.7m,

Specific Gravity S. G. of material = 1.21,

acceleration due to gravity in the Mars gm = 3.7m/s^2

We know that Weight W = mass m × acceleration due to gravity.

Let the Weight in on the Mars be Wm.

Wm = m × gm

Since we are given gm, we need to calculate for m. (Note that mass m is the same everywhere)

But mass = specific gravity × volume

Since we know the specific gravity, let's go ahead to calculate for the volume of the ball.

We know that Volume of a Sphere V = (4/3)πr^3

To get r, we know that C = 2πr

Therefore, r = C/(2π) = 0.7/(2π) = (7/10)/2π = 7/20π (in meters)

V = (4/3)*π×(7/20π)^3 = 343/6000π^2 (in meter^3)

m = 343/6000π^2 × 1.21 = 7.01×10^(-3)kg

Wm = 7.01×10^(-3) × 3.7 = 0.02593N

8 0

3 years ago

1. A piston having a diameter of 5.48 inches and a length of 9.50 in slides downward with a velocity, V, through a vertical pipe

const2013 [10]

Answer:

V = 0.00459 ft/s

Explanation:

Since the Piston is moving downwards with a constant velocity V, from the first Newton’s law we know that all vertical forces, must have zero resultant (their sum over vertical axis must equal to zero). Therefore, force that pulls the piston down, is equalized by force of viscous friction Fd= Fvf = 0.5lb (lb here is the pound-force unit). We will relate F ѵ f with τ and from that derive the equation for V.

Fѵf = τ . A

Where τ = µ. du/dy = µ . V/b , and A = π . D . l from this Follows:

Fѵf= (V. A .µ )/b  V= ( Fѵf .b )/(A.µ)

Placing all the known values in the equation ( remember to transform inches to feet, by multiplying inches values with the factor 1/12), we obtain :

ft2

V = ((0.5lb) . (0.002/12 ft))/(π . (5.48/12 ft) . (9.50/12 ft) . (0.016 lb.s/(ft^2 )))

V = 0.00459 ft/s

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3 0

3 years ago