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2 years ago

What landforms form at convergent boundaries where two oceanic plates collide?

2 answers:

6 0

They are formed when two plates collide, either crumpling up and forming mountains or pushing one of the plates under the other and back into the mantle to melt. Convergent boundaries form strong earthquakes, as well as volcanic mountains or islands, when the sinking oceanic plate melts.

Bad White [126]2 years ago

4 0

It forms mountains and sometimes islands.

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What is an example of a renewable resource?

alukav5142 [94]

A car A house A phone they all can be renewable

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2 years ago

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A car travels three-quarters of the way around a circle of radius 20.0 m in a time of 3.0 s at a constant speed. the initial vel

schepotkina [342]

20.3 divided by 3.0 will get u velocity and v times 3.0s

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2 years ago

A slender uniform rod 100.00 cm long is used as a meter stick. Two parallel axes that are perpendicular to the rod are considere

Nataliya [291]

Answer:

The correct answer is D $I_{30}$ / $I_{50}$ = 1.5

Explanation:

In this exercise the moment of inertia equation should be used

I = ∫ r² dm

In addition to the parallel axis theorem

I = $I_{cm}$ + M D²

Where $I_{cm}$ is the moment of the center of mass, M is the total mass of the body and D the distance from this point to the axis of interest

Let's apply these relationships to our problem, the center of mass of a uniform rod coincides with its geometric center, in this case the rod is 1 m long, so the center of mass is in

L = 100.00 cm (1m / 100 cm) = 1.0000 m

$x_{cm}$ = 50 cm = 0.50 m

Let's calculate the moment of inertia for this point, suppose the rod is on the x-axis and use the concept of linear density

λ = M / L = dm / dx

dm = λ dx

Let's replace in the moment of inertia equation

I = ∫ x² ( λ dx)

We integrate

I = λ x³ / 3

We evaluate between the lower limits x = -L/2 to the upper limit x = L/2

I = λ/3 [(L/2)³ - (-L/2)³] = lam/3 [L³/8 - (-L³ / 8)]

I = λ/3 L³/4

I = 1/12 λ L³

Let's replace the linear density with its value

I = 1/12 (M/L) L³

I = 1/12 M L²

Let's calculate with the given values

I = 1/12 M 1²

I = 1/12 M

This point is the center of mass of the rod

Icm = I = 1/12 M = 8.333 10-2 M

Now let's use the parallel axis theorem to calculate the moment of resection of the new axis, which is 0.30 m from one end, in this case the distance is

D = $x_{cm}$ - x

D = 0.50 - 0.30

D = 0.20 m

Let's calculate

$I_{30}$ = $I_{cm}$ + M D²

$I_{30}$ = 1/12 M + M 0.202

$I_{30}$ = M (1/12 + 0.04)

$I_{30}$ = M 0.123

To find the relationship between the two moments of inertia, divide the quantities

$I_{30}$ / $I_{50}$ = M 0.123 / (M 8.3 10-2)

$I_{30}$ / $I_{50}$ = 1.48

The correct answer is d 1.5

6 0

2 years ago

A 13-cm-diameter cd has a mass of 25 g . part a what is the cd's moment of inertia for rotation about a perpendicular axis throu

faltersainse [42]

<span>Answer: For a disc, the moment of inertia about the perpendicular axis through the center is given by 0.5MR^2. where M is the mass of the disc and R is the radius of the disc. For the axis through the edge, use parallel axis theorem. I = I(axis through center of mass) + M(distance between the axes)^2 = 0.5MR^2 + MR^2 (since the axis through center of mass is the axis through the center) = 1.5 MR^2</span>

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2 years ago

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A girl rolls a ball up an incline and allows it to re- turn to her. For the angle and ball involved, the acceleration of the bal

zalisa [80]

Answer:

3.28 m

3.28 s

Explanation:

We can adopt a system of reference with an axis along the incline, the origin being at the position of the girl and the positive X axis going up slope.

Then we know that the ball is subject to a constant acceleration of 0.25*g (2.45 m/s^2) pointing down slope. Since the acceleration is constant we can use the equation for constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

X0 = 0

V0 = 4 m/s

a = -2.45 m/s^2 (because the acceleration is down slope)

Then:

X(t) = 4*t - 1.22*t^2

And the equation for speed is:

V(t) = V0 + a * t

V(t) = 4 - 2.45 * t

If we equate this to zero we can find the moment where it stops and begins rolling down, that will be the highest point:

0 = 4 - 2.45 * t

4 = 2.45 * t

t = 1.63 s

Replacing that time on the position equation:

X(1.63) = 4 * 1.63 - 1.22 * 1.63^2 = 3.28 m

To find the time it will take to return we equate the position equation to zero:

0 = 4 * t - 1.22 * t^2

Since this is a quadratic equation it will have to answers, one will be the moment the ball was released (t = 0), the other will eb the moment when it returns:

0 = t * (4 - 1.22*t)

t1 = 0

0 = 4 - 1.22*t2

1.22 * t2 = 4

t2 = 3.28 s

7 0

2 years ago