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Vesnalui [34]
3 years ago
12

Which of the following processes most directly helps create soil from rocks?

Physics
2 answers:
Luda [366]3 years ago
7 0
A. Weathering most directly helps create soil from rocks.
irina [24]3 years ago
4 0
A. Weathering I think
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The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static fricti
Colt1911 [192]

This question is incomplete, the complete question;

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.

Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb

Answer:

- the magnitude of force at point A is 79.1033 lb

- since FA < FA_max; Ladder WILL NOT slip

Explanation:

Given that;

∑'MA = 0

⇒ NB [Lsin∅] - W[L/2.cos∅] = 0

NB = W / 2tan∅ -------let this be equation 1

∑Fx = 0

⇒ FA - NB = 0

FA = NB

therefore from equation 1

FA = NB = W / 2tan∅

we substitute in our values

FA = NB = 76 / 2tan(60°) = 21.9393 lb

Now ∑Fy = 0

NA - W = 0

NA = W = 76 lb

Net force at A will be

FA' = √( NA² + FA²)

= √( (W)² + (W / 2tan∅)²)

we substitute in our values

FA' = √( (76)² + (21.9393)²)

= √( 5776 + 481.3328)

= √ 6257.3328

FA' = 79.1033 lb

Therefore the magnitude of force at point A is 79.1033 lb

Now maximum possible frictional force at A

FA_max = μ × NA

so, FA_max = 0.4 × 76

FA_max = 30.4 lb

So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.

Therefore since FA < FA_max; Ladder WILL NOT slip

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2 years ago
The mere exposure effect is
gulaghasi [49]
DescriptionThe mere-exposure effect is a psychological phenomenon by which people tend to develop a preference for things merely because they are familiar with them.
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3 years ago
Competitive pressures stemming from the threat of entry are stronger when
navik [9.2K]

Answer:

Industries outlook is uncertain

Explanation:

Competitive pressures stemming from the threat of entry are stronger when the industry's outlook is uncertain or highly risky, entry barriers are low, and very few existing industry members are looking to expand their market reach by entering product segments or geographic areas where they currently do not have a presence. entry barriers are low, the pool of entry candidates is large, and existing industry members are earning good profits. there are fewer than 10 entry candidates with the potential to hurdle the industry's barriers to entry. t is difficult or costly for a customer to switch to a new brand, the total dollar investment needed to enter the market successfully exceeds $5 million, and existing governmental regulations impose significant cost and compliance burdens on industry members. buyers have strong brand preferences and high degrees of loyalty to their preferred brand and when it takes new entrants less than 5 years to secure attractive amounts of space on retailers' shelves and build a well-recognized brand name.

6 0
3 years ago
What is the speed of an object traveling a distance of 25 meters in 25 seconds
sergeinik [125]

Answer:

1 meter per second

Explanation:

Speed=Distance/Time

S=D/T

S=25/25

S=1

6 0
3 years ago
Read 2 more answers
Suppose we have two planets with the same mass, but the radius of the second one is twice the size of the first one. How does th
bogdanovich [222]

The free-fall acceleration on the second planet is one-fourth the value of the first planet.

Calculation:

Consider the mass of planet A to be, M

               the mass of planet B to be, Mₓ = M

               the radius of planet A to be, R₁

               the radius of planet B to be, R₂

The acceleration due to gravity on planet A's surface is given as:

g = GM/R₁²      - (1)

Similarly, the acceleration due to gravity on planet B's surface is given as:

g' = GM/R₂²                           [where, R₂ = 2R₁]

   = GM/4R₁²    -(2)

From equation 1 & 2, we get:

g/g' = GM/R₁² ÷ GM/4R₁²

g/g' = 4/1

Thus we get,

g' = 1/4 g

Therefore, the free-fall acceleration on the second planet is one-fourth the value of the first planet.

Learn more about free-fall here:

<u>brainly.com/question/13299152</u>

#SPJ4

6 0
2 years ago
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