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3 years ago

Island arcs are a product of _ whereas mountain ranges are a product of _. Volcanic

Physics

1 answer:

Nadusha1986 [10]3 years ago

4 0

Answer:

c) Subduction of an oceanic plate under another oceanic plate, collision of two continental plates, subduction of an oceanic plate under a continental plate

Explanation:

Islands arcs are a product of subduction of an oceanic plate under another oceanic plate

Mountain ranges are a product of collision of two continental

plates,

Volcanic arcs are a product of subduction of an oceanic plate under a continental plate

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An undiscovered planet, many lightyears from Earth, has one moon in a periodic orbit. This moon takes 23.0 days on average to co

Gelneren [198K]

Answer:

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Explanation:

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3 years ago

Part Two: Criteria, Constraints, and Prioritizations

AlexFokin [52]

Answer:

Usually, a solution can have several criteria and constraints. Even though all are important, some criteria are more important than others. The same holds true for constraints. But what do you do if it's impossible for a solution to cover every criterion while avoiding every constraint? In cases like this, you can use prioritization. Listing criteria and constraints based on priority shows the relative importance of each. You will need to prioritize the criteria and constraints for each sub-problem so that you can design a solution for each one individually. Prioritization can help you compare two different possible solutions. For example, the criterion that cars travel at 15 mph through the neighborhood might be a higher priority than the constraint that homeowners are only willing to spend $10,000 on this issue. If this is the case, you would want to generate solutions that also follow the priority in mind. All criteria are important, but engineers must sometimes make a trade-off, which is a compromise or change in one or more criteria or constraints so that they can be met at the same time. This is where prioritization comes in handy as it helps determine the trade-offs. A solution that is doing a better job of meeting one criterion may result in not completely meeting another criterion. Prioritization will help you choose which solution to go with.

Explanation:

I got this from quizlet :)

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2 years ago

A sphere of mass m and radius r is released from rest at the top of a curved track of height H. The sphere travels down the curv

iren2701 [21]

Explanation:

(a) On the dots below, which represent the sphere, draw and label the forces (not components) that are exerted on the sphere at point A and at point B, respectively. Each force must be represented by a distinct arrow starting on and pointing away from the dot.

At point A, there are three forces acting on the sphere: weight force mg pulling down, normal force N pushing left, and static friction force Fs pushing down.

At point B, there are three forces acting on the sphere: weight force mg pulling down, normal force N pushing down, and static friction force Fs pushing right.

(b) i. Derive an expression for the speed of the sphere at point A.

Energy is conserved:

PE = PE + KE + RE

mgH = mgR + ½mv² + ½Iω²

mgH = mgR + ½mv² + ½(⅖mr²)(v/r)²

mgH = mgR + ½mv² + ⅕mv²

gH = gR + ⁷/₁₀ v²

v² = 10g(H−R)/7

v = √(10g(H−R)/7)

ii. Derive an expression for the normal force the track exerts on the sphere at point A.

Sum of forces in the radial (-x) direction:

∑F = ma

N = mv²/R

N = m (10g(H−R)/7) / R

N = 10mg(H−R)/(7R)

(c) Calculate the ratio of the rotational kinetic energy to the translational kinetic energy of the sphere at point A.

RE / KE

= (½Iω²) / (½mv²)

= ½(⅖mr²)(v/r)² / (½mv²)

= (⅕mv²) / (½mv²)

= ⅕ / ½

= ⅖

(d) The minimum release height necessary for the sphere to travel around the loop and not lose contact with the loop at point B is Hmin. The sphere is replaced with a hoop of the same mass and radius. Will the value of Hmin increase, decrease, or stay the same? Justify your answer.

When the sphere or hoop just begins to lose contact with the loop at point B, the normal force is 0. Sum of forces in the radial (-y) direction:

∑F = ma

mg = mv²/R

gR = v²

Applying conservation of energy:

PE = PE + KE + RE

mgH = mg(2R) + ½mv² + ½Iω²

mgH = 2mgR + ½mv² + ½(kmr²)(v/r)²

mgH = 2mgR + ½mv² + ½kmv²

gH = 2gR + ½v² + ½kv²

gH = 2gR + ½v² (1 + k)

Substituting for v²:

gH = 2gR + ½(gR) (1 + k)

H = 2R + ½R (1 + k)

H = ½R (4 + 1 + k)

H = ½R (5 + k)

For a sphere, k = 2/5. For a hoop, k = 1. As k increases, H increases.

(e) The sphere is again released from a known height H and eventually leaves the track at point C, which is a height R above the bottom of the loop, as shown in the figure above. The track makes an angle of θ above the horizontal at point C. Express your answer in part (e) in terms of m, r, H, R, θ, and physical constants, as appropriate. Calculate the maximum height above the bottom of the loop that the sphere will reach.

C is at the same height as A, so we can use our answer from part (b) to write an equation for the initial velocity at C.

v₀ = √(10g(H−R)/7)

The vertical component of this initial velocity is v₀ sin θ. At the maximum height, the vertical velocity is 0. During this time, the sphere is in free fall. The maximum height reached is therefore:

v² = v₀² + 2aΔx

0² = (√(10g(H−R)/7) sin θ)² + 2(-g)(h − R)

0 = 10g(H−R)/7 sin²θ − 2g(h − R)

2g(h − R) = 10g(H−R)/7 sin²θ

h − R = 5(H−R)/7 sin²θ

h = R + ⁵/₇(H−R)sin²θ

4 0

3 years ago

A compressed spring is compressed between a 5kg and 3 kg cart, respectively. When the spring is released, what will be the same

stira [4]

The momentum of the two carts will be the same

Explanation:

First of all, we notice that the system consisting of the two carts + the spring is an isolated system: this means that there are no external forces acting on it.

As a result, the total momentum of the system must be conserved.

Since the spring is at rest, the total momentum of the system after the two carts are released is:

$p_f = p_1 + p_2$