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2 years ago

For a given sand soil, the maximum and minimum void ratios are 0.78 and 0.43, respectively.

Geography

1 answer:

Zinaida [17]2 years ago

4 0

Answer:

hope this helped

Explanation:

a.) Dry unit weight b.) Void ratio c.) Porosity d.) Degree of saturation Solution: a) γd= γ1+?= 17.611.108= 15.89 kN/m3b) γd= γ? ?1+15.89 = (2.67)(9.81)1+e = 0.65 c) n = e1+= 0.651.65= 0.39 d) Se = wGsS = (0.108)(2.67)0.65x 100% = 44.50% 3.2Refer to Problem 3.1. Determine the weight of water, in KN, to be added per cubic meter of soil for: a.) 80% DEGREE OF SATURATION b.) 100% DEGREE OF SATURATION Solution: a. ) S = v? ??∆S = 80 –44.50 = 35.5% n = v? ?; VT = 1m3 0.39 = v? 1Vv = 0.39 0.355 = v? 0.39; Vw = (0.355)(0.39) Vw = 0.140 m3 Ww = 0.140 x 9.81 = 1.37 Kn

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A radioactive isotope of the element potassium decays to produce argon. If the ratio of argon to potassium in a mineral is found

Shkiper50 [21]

Answer:

After three half-lives, it is 7:1

Explanation:

The rate at which the decay of the radioactive isotope is said to be half-life and is represented as the half-life of the atoms and the matter to disintegrate itself.

The two radioactive isotopes are K 40 and Ar 40. After the K-Ar dating, they contain minerals such as the clay mica and the temphra minerals and Half-life of the K 40 is 1.3 more than 100,000 years i.e 10x9 years for Ar 40.

<u>Thus the K 40 has to divide as three half-lives have passed if the ratio of argon to potassium atoms is 7:1. At the end of one half-life, the ratio of the number of argon to potassium atoms is 1:1. and two half lives its 3:1 thus after the third half-life it becomes as 7:1.</u>

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2 years ago

sergejj [24]

Answer:

Explanation:

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2 years ago

5. How fast did the shoes travel? (4 points) a. From where the shoes spilled (48°N, 161°W) to where they first made landfall (49

Leto [7]

Answer:

a) d = 35.0143 Km

t = 188 days

v = 7.76*10⁻³ Km/h

b) d = 28.6356 Km

t = 2165 days

v = 5.51*10-4 Km/h

Explanation:

a. From where the shoes spilled (48°N, 161°W) to where they first made landfall (49°N, 126°W), how many kilometers did they travel?

We can use the equation

d = √(∆N²+∆W²)

⇒ d = √((49-48)²+(126-161)²) = 35.0143 Km

How many days did they take to travel that distance? (You can use November 30 as the date found).

From May 27, 1990 to November 30, 1990 we have

t = 5 days + (30 days/month)*(3 months) + (31 days/month)*(3 months) = 188 days

⇒ t = 188 days = (188 days)*(24 hours/day) = 4512 hours

What was their rate of travel in kilometers per hour?

We use the equation

v = d/t ⇒ v = 35.0143 Km/ 4512 h

⇒ v = 7.76*10⁻³ Km/h

b. From where the shoes spilled (48°N, 161°W) to where they were found in 1996 (54°N, 133°W), how many kilometers did they travel?

We can use the equation

d = √(∆N²+∆W²)

⇒ d = √((54-48)²+(133-161)²) = 28.6356 Km

How many days did they take to travel that distance (use April 30 as the date found)?

From May 27, 1990 to April 30, 1996 we have

t = 5 days + (30 days/month)*(4 months) + (31 days/month)*(6 months) + 29 days + (365 days/ year)*(4 years)+ (366 days/ year)*(1 year)= 2165 days

⇒ t = 2165 days = (2165 days)*(24 hours/day) = 51960 hours

What was their rate of travel in kilometers per hour?

We use the equation

v = d/t ⇒ v = 28.6356 Km/ 51960 h

⇒ v = 5.51*10-4 Km/h

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