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Ivenika [448]
3 years ago
7

Digging downward through a soil profile, how deep would you have to go before you found recognizable fragments of the parent roc

k for the soil?
Physics
1 answer:
ohaa [14]3 years ago
3 0
You Willis have to go about 6 -7 ft down
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Explain about kinetic theory​
Luden [163]

Answer:

The model, called the kinetic theory of gases, assumes that the molecules are very small relative to the distance between molecules. ... The molecules are in constant random motion, and there is an energy (mass x square of the velocity) associated with that motion. The higher the temperature, the greater the motion.

6 0
3 years ago
I will mark brainliest!! The Moon travels around the Earth each month in an elliptical path, as opposed to a perfect circle.
Natasha_Volkova [10]

Answer:

The magnitude decreases

Explanation:

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2 years ago
What is the density of 18.0-karat gold that is a mixture of 18 parts gold, 5 parts silver, and 1 part copper? (These values are
nexus9112 [7]

Answer:

Density of 18.0-karat gold mixture is 15.58 g/cm^3.

Explanation:

A mixture of 18 parts gold, 5 parts silver, and 1 part copper.

Let mass of gold be 18x

Let the mass of silver be 5x

Let the mass of copper be 1x

The density of gold = 19.32g/cm^3

The density of silver = 10.1g/cm^3

The density of copper =8.8g/cm^3

Volume=\frac{Mass}{Density}

Volume of the gold in the mixture = V_1=\frac{18x}{19.32 g/cm^3}

Volume of the silver in the mixture = V_2=\frac{5x}{10.1 g/cm^3}

Volume of the copper in the mixture = V_3=\frac{1x}{8.8 g/cm^3}

Mass of the mixture = M = 18x+5x+1x =24x

Volume of the mixture = V_1+V_2+V_3

Density of the mixture:

\frac{M}{V_1+V_2+V_3}=15.58 g/cm^3

8 0
3 years ago
Help please!!!!!!!!!!!!!!!!!!!!
Pavel [41]
Looks like you need to review through the lesson and take notes as it tells you in the lesson what each of these are.
8 0
3 years ago
You run<br> completely around a 400m track in<br> 80s. What was your average velocity?
dalvyx [7]

Answer:

V=?

S=400m

t=80s

V=S/t

V=400/80

V=5m/sec

6 0
3 years ago
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