What would be the required voltage of an energy source in a circuit with a current of 10.0 A and a resistance of 11.0 Ω?

2. The moist weight of 0.1 ft3 of soil is 12.2 lb. If the moisture content is 12% and the specific gravity of soil solids is 2.7

adell [148]

The answers to dry unit weight, void ratio, porosity, degree of saturation, volume occupied by water are respectively;

γ_d = 108.93 lb/ft³; e = 0.56; n = 0.36; S = 0.58; V_w = 0.021 ft³

<h3>Calculation of Volume and Weight of soil</h3>

We are given;

Moist weight; W = 12.2 lb

Volume of moist soil; V = 0.1 ft³

moisture content; w = 12% = 0.12

Specific gravity of soil solids; G_s = 2.72

A) Formula for dry unit weight is;

γ_d = γ/(1 + w)

where γ_w is moist unit weight as;

γ_w = W/V

γ_w = 122/0.1 = 122 lb/ft³

Thus;

γ_d = 122/(1 + 0.12)

γ_d = 108.93 lb/ft³

B) Formula for void ratio is;

e = [(G_s * γ_w)/γ_d] - 1

e = [(2.72 * 122)/108.93] - 1

e = 0.56

C) Formula for porosity is;

n = e/(1 + e)

n = 0.56/(1 + 0.56)

n = 0.36

D) Formula for degree of saturation is;

S = (w * G_s)/e

S = (0.12 * 2.72)/0.56

S = 0.58

E) Volume occupied by water is gotten from;

V_w = S*V_v

where;

V_v is volume of voids = nV

V_v = 0.36*0.1

V_v = 0.036 ft³

Thus;

V_w = 0.58 * 0.036

V_w = 0.021 ft³

Read more about Specific Gravity of Soil at; brainly.com/question/14932758

4 0

3 years ago

A steel pipe of 400-mm outer diameter is fabricated from 10-mm-thick plate by welding along a helix that forms an angle of 20° w

Verdich [7]

Explanation:

Outer di ameter $d_{0}=400 \mathrm{mm}[tex] Thickness of the cylinder [tex]t=10 \mathrm{mm}$

$\therefore[tex] Inner diam eter [tex]d_{i}=d_{0}-2 t=400-2 \times 10$

$d_{1}=380 \mathrm{mm}$

Given loading on the cylinder $P=300 \mathrm{kN}$ Helix an gle of the weld form $\theta=20^{\circ}$

(i) Normal stress on the plane at angle $\theta=20^{\circ}$ is

$\sigma=\frac{P \cos ^{2} \theta}{A_{0}}$

$\text { Where } A_{0}=\frac{\pi}{4}\left(d_{0}^{2}-d_{1}^{2}\right)$

$\quad=\frac{\pi}{4}\left(400^{2}-380^{2}\right)$

$=12252.21 \mathrm{mm}^{2}$

$=12.25221 \times 10^{-9} \mathrm{m}^{2}$

$\sigma=\frac{-300 \times 10^{2} \times \cos ^{2} 20}{12.25221 \times 10^{-1}}$

$=-21.6 \mathrm{MPa}$

(ii) Shear stress along an angle of $\theta=20^{\circ}$ is $\tau=\frac{P}{A_{0}} \cos \theta \sin \theta$

$=\frac{-300 \times 10^{-1} \times \cos 20 \times \sin 20}{12.25221 \times 10^{-3}}$

$=-7.86 \mathrm{MPa}$

3 0

3 years ago

If 9 + 10 = 21, then what is 10 + 10?

Katen [24]

The answer would be 23 because the engines in a horse is a amount of 9 engines in 69 cars

6 0

2 years ago

Read 2 more answers

Which of these materials are insulators? Select the THREE (3) that apply.

lubasha [3.4K]

Answer:

wool, rubber, and plastic

Explanation:

5 0

3 years ago

Explain why failure of this garden hose occurred near its end and why the tear occurred along its length. Use numerical values t

alukav5142 [94]

Answer:

hoop stress
longitudinal stress
material used

all this could led to the failure of the garden hose and the tear along the length

Explanation:

For the flow of water to occur in any equipment, water has to flow from a high pressure to a low pressure. considering the pipe, water is flowing at a constant pressure of 30 psi inside the pipe which is assumed to be higher than the allowable operating pressure of the pipe. but the greatest change in pressure will occur at the end of the hose because at that point the water is trying to leave the hose into the atmosphere, therefore the great change in pressure along the length of the hose closest to the end of the hose will cause a tear there. also the other factors that might lead to the failure of the garden hose includes :

hoop stress ( which acts along the circumference of the pipe):

αh = $\frac{PD}{2T}$ EQUATION 1