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3 years ago

PART ONE

Physics

1 answer:

stepladder [879]3 years ago

5 0

Answer:

1. A1, B2, C3

2. 47.1°

Explanation:

Sum of forces in the x direction:

∑Fₓ = ma

f − Fᵥᵥ = 0

f = Fᵥᵥ

Sum of forces in the y direction:

∑Fᵧ = ma

N − W = 0

N = W

Sum of moments about the base of the ladder:

∑τ = Iα

Fᵥᵥ h − W (b/2) = 0

Fᵥᵥ h = ½ W b

Fᵥᵥ (l sin θ) = ½ W (l cos θ)

l Fᵥᵥ sin θ = ½ l W cos θ

The correct set of equations is A1, B2, C3.

At the smallest angle θ, f = Nμ. Substituting into the first equation, we get:

Nμ = Fᵥᵥ

Substituting the second equation into this equation, we get:

Wμ = Fᵥᵥ

Substituting this into the third equation, we get:

l (Wμ) sin θ = ½ l W cos θ

μ sin θ = ½ cos θ

tan θ = 1 / (2μ)

θ = atan(1 / (2μ))

θ = atan(1 / (2 × 0.464))

θ ≈ 47.1°

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A scientist observes rock masses that have moved past each other in opposite horizontal directions. Which feature

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Answer:

C. strike-slip fault

Explanation:

The scientist must have observed a strike- slip fault.

A fault is an evidence of brittle deformation of the crust in the presence of applied stress on earth materials. Here, the earth material is the rock subjected to tension.

Where a fault occurs, there must have been movement between two blocks of rocks. The direction of movement helps us to delineate the fault type.

When two blocks moves past each other horizontally, it is a strike-slip fault like rubbing your palms together.
When a block moves in the direction of the dip, it forms a dip-slip fault which results in a fault-block mountain characterized by graben and horst systems.

Option A, Plateau is a table landform usually a mountain with flat peak.

Option B is a bowl shaped stratigraphic pattern in which the youngest sequence is at the core of the strata or a fold.

So, the most fitting option is C, a strike-slip fault.

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3 years ago

Which of the following woud be expected to have the greatest impact on primary productivity?

leonid [27]

Answer: C solar energy

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Explanation:

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3 years ago

We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top

notka56 [123]

Answer:

A,)FD= 114.1N

B)Torque=798.5Nm

Explanation:

We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top of the tree exerts a horizontal force, and thus a torque that can topple the tree if there is no opposing torque. Suppose a tree's canopy presents an area of 9.0 m^2 to the wind centered at a height of 7.0 m above the ground. (These are reasonable values for forest trees.)

If the wind blows at 6.5 m/s, what is the magnitude of the drag force of the wind on the canopy? Assume a drag coefficient of 0.50 and the density of air of 1.2 kg/m^3

B)What torque does this force exert on the tree, measured about the point where the trunk meets the ground?

A)The equation of Drag force equation can be expressed below,

FD =[ CD × A × ρ × (v^2/ 2)]

Where CD= Drag coefficient for cone-shape = 0.5

ρ = Density

Area of of the tree canopy = 9.0 m^2

density of air of = 1.2 kg/m^3

V= wind velocity= 6.5 m/s,

If we substitute those values to the equation, we have;

FD =[ CD × A × ρ × (v^2/ 2)]

F= [ 0.5 × 9.0 m^2 × 1.2 kg/m^3 ( 6.5 m/s/ 2)]

FD= 114.1N

B) the torque can be calculated using below formula below

Torque= (Force × distance)

= 114.1 × 7

= 798.5Nm

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3 years ago

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Answer:

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Explanation:

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3 years ago

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7. Two people are pushing a 40.0kg table across the floor. Person 1 pushes with a force of 490N

artcher [175]

Answer:

20.4 $m/s^{2}$

Explanation:

To start doing this problem, first draw a free body diagram of the table. My teacher always tells us to do this, and I find that it is very helpful. I have attached a free body diagram to this answer- take a look at it.

First, let us see if Net force = MA. To do that, we need to determine whether the object is at equilibrium horizontally. For an object to be at equilibrium, it either needs to be moving at a constant velocity or not moving at all. Also, if an object is at equilibrium, there will not be any acceleration. But we know that there IS acceleration horizontally, so it cannot be in equilibrium. If it is not in equilibrium, we can use the formula ∑F= ma.

Let us determine the net force. Since the object is moving horizontally, we can ignore the weight and normal force, because they are vertical forces. The only horizontal forces we need to worry about are the applied force and force of friction.

Applied force = 1055 N (490 + 565)

Friction force= Unknown

To find the friction force, use the kinetic friction formula, Friction = μkN

μk is the coefficient, which the problem includes- it is 0.613.

N is the normal force, which we have to find.

*To find the normal force, we have to determine if the object is at equilibrium VERTICALLY. Since it has no acceleration vertically (it's not moving up/down), it is at equilibrium. Now, when an object is at equilibrium in one direction, it means that all the forces in that direction are equal. What are our vertical forces? Weight (mg) and Normal force (N). So it means that the Normal force is equal to the Weight.

Weight = mg = (40)(9.8) = 392 N

Normal force = 392 N

Now, plug it back into the formula (μkN): (0.613)(392) = 240.296 N

Friction = 240.296 N

Now that we know the friction, we can find the horizontal net force. Just subtract the friction force, 240.296 from the applied force, 1055 N

Horizontal Net Force: 814.704 N

Now that we know the net force, plug in the numbers for the formula

∑F= ma.

814.704 = (40.0)(a)

*Divide on both sides)

a = 20.3676 m/s^2

Round it to 3 significant figures, to get:

20.4 $m/s^{2}$

7 0

3 years ago