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barxatty [35]
3 years ago
11

When a volcano erupts, tiny particles from which of Earth’s spheres are released into the air?

Physics
2 answers:
prisoha [69]3 years ago
6 0

Answer

Option C: geosphere

Explanation

♡ ∩_∩

(„• ֊ •„)♡

┏━∪∪━━━━┓

♡ good luck 。 ♡

┗━━━━━━━┛

SSSSS [86.1K]3 years ago
3 0

Answer:Atmosphere

When a volcano erupts, particles of rock and ash are released into the atmosphere. After this, water droplets form around the rock and ash particles and fall to Earth as rain.

Explanation:

Hope it helps!

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A tectonic plate near Niihau, an island in Hawaii, grows at an average rate of 11 cm/year. If the rate of plate expansion double
RideAnS [48]

Answer:

C, 66

Explanation:

11*2 is 22 and 22*3 is 66

5 0
4 years ago
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A gyroscope flywheel of radius 1.96 cm is accelerated from rest at 13.0 rad/s2 until its angular speed is 2270 rev/min. (a) What
nika2105 [10]

Tangential acceleration of a point on the rim of the flywheel during this spin-up process is 0.2548 m/s².

Tangential acceleration is defined as the rate of change of tangential velocity of the matter in the circular path.

Given,

Radius of flywheel (r) = 1.96 cm = 0.0196m

Angular acceleration (α)= 13.0 rad/s²

The tangential acceleration formula is at=rα

where, α is the angular acceleration, and r is the radius of the circle.

using the formula; at=rα  = (13.0 rad/s²) (0.0196m) = 0.2548 m/s².

The tangential acceleration is 0.2548 m/s².

Learn more about the Tangential acceleration with the help of the following link:

brainly.com/question/15743294

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5 0
2 years ago
Conduction occurs when thermal energy is transferred by the movement of
Alex

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fluids

Explanation:

5 0
3 years ago
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A playground merry-go-round of radius R = 2.20 m has a moment of inertia I = 260 kg · m2 and is rotating at 12.0 rev/min about a
MrRa [10]

Answer:

The new angular speed of the merry-go-round is 8.31 rev/min.

Explanation:

Because the merry-go-round is rotating about a frictionless axis there’re not external torques if we consider the system merry-go-round and child. Due that we can apply conservation fo angular momentum that states initial angular momentum (Li) should be equal final angular momentum (Lf):

L_f=L_i (1)

The initial angular momentum is just the angular momentum of the merry-go-round (Lmi) that because it's a rigid body is defined as:

L_i=L_{mi}=I\omega_i (2)

with I the moment of inertia and ωi the initial angular speed of the merry-go-round

The final angular momentum is the sum of the final angular momentum of the merry-go-round plus the final angular momentum of the child (Lcf):

L_f=L_{mf}+L{cf}=I\omega_f+L{cf} (3)

The angular momentum of the child should be modeled as the angular momentum of a punctual particle moving around an axis of rotation, this is:

L{cf}=mRv_f (4)

with m the mass of the child, R the distance from the axis of rotation and vf is final tangential speed, tangential speed is:

v_f=\omega_f R (5)

(note that the angular speed is the same as the merry-go-round)

using (5) on (4), and (4) on (3):

L_f=I\omega_f+m\omega_f R^2 (6)

By (5) and (2) on (1):

I\omega_f+m\omega_f R^2=I\omega_i

Solving for ωf (12.0 rev/min = 1.26 rad/s):

\omega_f= \frac{I\omega_i}{]I+mR^2}=\frac{(260)(1.26)}{260+(24.0)(2.20)^2}

\omega_f=0.87\frac{rad}{s}=8.31 \frac{rev}{min}

8 0
3 years ago
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