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saveliy_v [14]
3 years ago
14

A rock in a harsh desert is slowly breaking down over time. Temperatures in the desert rise quickly during the day and fall quic

kly at night. These temperature changes cause stress in the rock, and as a result, it is slowly breaking apart.
Which of the following processes is happening to this rock?
Geography
2 answers:
Reil [10]3 years ago
7 0

Answer:b. physical weathering

Explanation:

Goshia [24]3 years ago
5 0

Answer: Physical/Mechanical Weathering

Explanation:

Weathering is the process that breaks down rocks into smaller particles like sediment and soil. Water, ice, acids, salts, plants, animals, and changes in temperature are all agents of weathering. In this case, extreme heat during the day may have caused the rock to expand, only to shrink from the cold of the night. Over time, this weakened the structure of the rock, causing it to split and crumble, with outer layers flaking off in the process known as exfoliation.

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Solve the equation<br>p-(p+20) = 4(p-5)+ 3p<br>The solution is p =<br>I​
EleoNora [17]
Answer: See below

Explanation:
p-(p+20) = 4(p-5) + 3p
p-p-20 = 4p - 20 + 3p
-20 = 7p -20
40 = 7p
40/7 = p
8 0
3 years ago
NWT ooking for. the unknown number (X) the dimensions Clength and uwidth) What do I know? length 1 cm less than (1= 2x+1) turce
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That is the wrong subject
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3 years ago
Could you please help me on this question it is very difficulted geometry question.
N76 [4]

Answer:

21.5%

Explanation:

to find the answer to this equation you must found out how much each of the areas are

the circles area is represented by pi*(5^2)

that is 25pi

or

25 * pi = 78.5398163

then you find the area of the square

10 * 10 = 100 (5 + 5 = 10)

you can directly subtract 78.5398163 from 100 because it can be portrayed as 100%

100 - 78.5398163 = 21.4601837

you must round to the nearest tenth so your answer is

21.5%

6 0
2 years ago
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3 0
3 years ago
To measure a base lone, a steel stape 30m lang, Standardised at 15°C with a pull of 80N was used. find the correction per tape l
iren2701 [21]

The correction of thermal expansion per tape length, if the temperature at the time of measurements is 25°c and the pull exerted is 150N is 4.67 ×10 ⁻³m.

Thermal expansion is the tendency of remembering to change in form, volume, and region in response to an alternate in temperature. Temperature is a monotonic characteristic of the common molecular kinetic power of a substance.

Given,

Length of steel tape = 30m

Temperature at standardization = 15°C

The temperature at the time of measurement = 25°C

Standardized pull (Po) = 80N

Pull exerted (Pm) = 150N

Young modulus of elasticity = E =  E= 2X10^5 N/mm2

Coefficient of thermal expansion = 211.2x10^6c°

Cross-sectional area of tape = 8 mm²

For temperature correction  Cm = (Tm - T₀)αL₀

Putting α = ( 25 - 15 ) × 11.2 × 10⁻⁶ × 30

            Cm = 3.36 ×  10⁻³ m

For full correction ;

         C pull = (Pm - Po ) × Lo

Putting value'

    C pull =  (150 - 80) × 30 ×  1000 /8 ×2 × 10⁵

    C pull = 1.31 × 10 ⁻³

Total correction = 3.36 × 10⁻³ m + 1.31 × 10 ⁻³

Total correction = 4.67 ×10 ⁻³

Learn more about thermal expansion here:-brainly.com/question/1166774

#SPJ9

8 0
2 years ago
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