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Yuri [45]
2 years ago
9

Precast concrete curtain wall panels: Group of answer choices are typically manufactured on site and then hoisted into place. of

ten are unreinforced.
Engineering
1 answer:
amid [387]2 years ago
5 0

Answer:

c) can be made with a variety of surface finishes.

Explanation:

The missing options are;

When it comes to concrete work in construction, the concrete can be cast either in-situ or in form of pre-cast concrete.

Now in-situ concrete means concrete done on the construction site being built while pre cast concrete simply means concrete cast outside in a factory or yard and brought to site to mount.

These pre cast concrete could have different surface finishes as required as this is one of it's advantages over in situ because there is a lot of space and room to have the desired concrete finish.

a) are typically manufactured on site and then hoisted into place.

b) cannot be fiber-reinforced.

c) can be made with a variety of surface finishes.

d) never include insulation.

e) often are unreinforced.

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A heat engine is coupled with a dynamometer. The length of the load arm is 900 mm. The spring balance reading is 16. Applied wei
miss Akunina [59]

Answer:

P = 80.922 KW

Explanation:

Given data;

Length of load arm is 900 mm = 0.9 m

Spring balanced  read 16 N

Applied weight is 500 N

Rotational speed is 1774 rpm

we know that power is given as

P = T\times \omega

T Torque = (w -s) L = (500 - 16)0.9 = 435.6 Nm

\omega angular speed =\frac{2 \pi N}{60}

Therefore Power is

P =\frac{435.6 \time 2 \pi \times 1774}{60} = 80922.65  watt

P = 80.922 KW

4 0
3 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-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
3 years ago
determine the optimum compressor pressure ratio specific thrust fuel comsumption 2.1 220k 1700k 42000 1.004
Afina-wow [57]

Answer:

hello your question is incomplete attached below is the complete question

A) optimum compressor ratio = 9.144

B) specific thrust = 2.155 N.s /kg

C) Thrust specific fuel consumption = 1670.4 kg/N.h

Explanation:

Given data :

Mo = 2.1 ,  To = 220k , Tt4 = 1700 k, hpr = 42000 kj/kg, Cp = 1.004 kj/ kg.k

γ = 1.4

attached below is the detailed solution

6 0
3 years ago
As the asteroid falls closer to the Earth's surface its _______ energy decreases and its _______ energy increases.
Levart [38]

Answer:

As the asteroid falls closer to the Earth's surface its <u>Gravitational</u> <u>Potential</u> energy <em>decreases</em> and its <u>Kinetic</u> energy <em>increases</em>.

4 0
3 years ago
Part A Engineering stress and strain are calculated using the actual cross-sectional area and length of the specimen. True or fa
Galina-37 [17]

Answer: True

Explanation:

Engineering stress is the applied load divided by the original cross-sectional area of a material. It is also known as nominal stress. It can also be defined as the force per unit area of a material. Engineering Stress is usually in large numbers.

While Engineering strain is the amount that a material deforms per unit length in a tensile test.  It can also be defined as extension per unit length. It has no unit as it is a ratio of lengths. Engineering Strain is in small numbers.

5 0
3 years ago
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