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Mashcka [7]
3 years ago
15

Study the image. Point 1 is beneath water moving toward land. Point 2 is next to an erupting vocano. Point 3 is beneath Earth ap

ostrophe s crust moving toward the water. Point 4 is the same layer as Point 1 moving down below Point 3. At which point is subduction occurring? 1 2 3 4
Physics
2 answers:
juin [17]3 years ago
7 0

Answer:

I believe the answer is D) 4

Explanation:

Subduction is the sideways and downward movement of the edge of a plate of the earth's crust into the mantle beneath another plate.

svet-max [94.6K]3 years ago
5 0

Answer:

c

Explanation:

mart me as brailist

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How much kinetic energy does a 55 kg object have that is traveling 20m/s?
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Answer:

11000 J

Explanation:

Apply the formula: K = 1/2.m.v²

K = 1/2.55.20²

K = 1/2.55.400

K = 55.200

K = 11000 Joules

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An 7.9 kg crate is pulled 5.5 m up a 30∘ incline by a rope angled 19 ∘ above the incline. The tension in the rope is 140 N and t
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Answer:

Wt= 0.73kJ, Wg=0.21kJ Wn= 0J

(B) Wf= 0.096kJ

Explanation:

See attachment below.

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Who revolves around who in the Earth, Sun and Moon system
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A solid wood door 1.00 m wide and 2.00 m high is hinged along one side and has a total mass of 40.0 kg. Initially open and at re
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Answer:

The final angular speed is 0.223 rad/s

Explanation:

By the conservation of angular moment:

ΔL=0

L₁=L₂

L₁ is the initial angular moment

L₂ is the final angular moment

L₁ is given by:

L_1=L_{door} + L_{mud}

As the door is at rest its angular moment is zero and the angular moment of mud can be considered as a point object, then:

L_1= L_{mud}= mvr

where

r is the distance from the support point to the axis of rotation (the mud hits at the center of the door; r=0.5 m)

v is the speed

m is the mass of the mud

L₂ is given by:

L_2= (I_{door} + I_{mud}) \omega_f

ωf is the final angular speed

The moment of inertia of the door can be considered as a rectangular plate:

I_{door}=\frac{1}{3}MW^2

M is the mass of the door

W is the width of the door

The moment of inertia of the mud is:

I_{mud}=mr^2

Hence,

L_1=L_2\\mvr= (I_{door} + I_{mud}) \omega_f\\\omega_f=\frac{mvr}{I_{door} + I_{mud}} \\\omega_f=\frac{mvr}{I_{door} + I_{mud}}

\omega_f=\frac{0.5kg \times 12m/s \times 0.5m}{\frac{1}{3}40kg(1m)^2+0.5kg \times (0.5m)^2}

\omega_f=0.223 \frac{rad}{s}

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4 years ago
If two objects have the same volume but one has a greater mass, the one with greater mass
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