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SVEN [57.7K]
3 years ago
14

The continental shelf, continental slope, and continental rise combine to form the

Physics
2 answers:
Feliz [49]3 years ago
8 0

Answer;

Continental margin

Explanation;

-Continental margin is the section of the ocean is composed of the continental shelf, the continental slope, and the continental rise.

-The continental margin is that portion of the ocean that separates the continents from the deep ocean floor. The continental margin is usually subdivided into three major sections: the continental shelf, the continental slope, and the continental rise. In addition to these sections, one of the most important features of the continental margin is the presence of very large submarine canyons that cut their way through the continental slope and, less commonly, the continental shelf.

BartSMP [9]3 years ago
4 0
The continental shelf, continental slope, and continental rise combine to form the <span>continental margin</span><span>.

</span>The continental shelf is the portion of the land mass that extends into ocean water.

Continental slope begins where the continental shelf ends. It does exactly what its name implies and serves as a boundary between the oceanic crust and the continental crust.

The continental slope divides the continental and oceanic crusts. Over a period of time, soil, rocks, and debris wash down the steep sides of a continental slope due to the influence of gravity. They pile up at the bottom of the slope and form a small ridge called continental rise.
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The flywheel of a steam engine runs with a constant angular velocity of 150 rev/min. When steam is shut off, the friction of the
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Answer:

a) -1.14 rev/min²

b) 9900 rev

c) -9.92×10⁻⁴ m/s²

d) 30.8 m/s²

Explanation:

First, convert hours to minutes:

2.2 h × 60 min/h = 132 min

a) Angular acceleration is change in angular velocity over change in time.

α = (ω − ω₀) / t

α = (0 rev/min − 150 rev/min) / 132 min

α = -1.14 rev/min²

b) θ = θ₀ + ω₀ t + ½ αt²

θ = 0 rev + (150 rev/min) (132 min) + ½ (-1.14 rev/min²) (132 min)²

θ = 9900 rev

c) The tangential component of linear acceleration is:

a_t = αr

First,  convert α from rev/min² to rad/s²:

-1.14 rev/min² × (2π rad/rev) × (1 min / 60 s)² = -1.98×10⁻³ rad/s²

Therefore:

a_t = (-1.98×10⁻³ rad/s²) (0.50 m)

a_t = -9.92×10⁻⁴ m/s²

d) The magnitude of the net linear acceleration can be found from the tangential component and the radial component:

a² = (a_t)² + (a_r)²

The radial component is the centripetal acceleration:

a_r = v² / r

a_r = ω² r

First, convert 75 rev/min to rad/s:

75 rev/min × (2π rad/rev) × (1 min / 60 s) = 7.85 rad/s

Find the radial component:

a_r = (7.85 rad/s)² (0.50 m)

a_r = 30.8 m/s²

Now find the net linear acceleration:

a² = (-9.92×10⁻⁴ m/s²² + (30.8 m/s²)²

a = 30.8 m/s²

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3 years ago
What is the oscillation period of your eardrum when you are listening to the A4 note on a piano (frequency 440 Hz)?
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T=\dfrac{1}{f}=\dfrac{1}{440}=0.0022\overline{72}

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3 years ago
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n alpha particle (q = +2e, m = 4.00 u) travels in a circular path of radius 5.94 cm in a uniform magnetic field with B = 1.10 T.
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Answer:

a). V = 3.13*10⁶ m/s

b). T = 1.19*10^-7s

c). K.E = 2.04*10⁵

d). V = 1.02*10⁵V

Explanation:

q = +2e

M = 4.0u

r = 5.94cm = 0.0594m

B = 1.10T

1u = 1.67 * 10^-27kg

M = 4.0 * 1.67*10^-27 = 6.68*10^-27kg

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Mv / r = qB

V = qBr / m

V = [(2 * 1.60*10^-19) * 1.10 * 0.0594] / 6.68*10^-27

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b). Period of revolution.

T = 2Πr / v

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T = 1.19*10⁻⁷s

c). kinetic energy = ½mv²

K.E = ½ * 6.68*10^-27 * (3.13*10⁶)²

K.E = 3.27*10^-14J

1ev = 1.60*10^-19J

xeV = 3.27*10^-14J

X = 2.04*10⁵eV

K.E = 2.04*10⁵eV

d). K.E = qV

V = K / q

V = 2.04*10⁵ / (2eV).....2e-

V = 1.02*10⁵V

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Kinetic energy is energy of motion. Pick choice-A, at the top of the swing,  where she stops moving & then goes the other way.

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A model train traveling at a constant speed around a circular track has a constant velocity
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The answer should be yes
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