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Pachacha [2.7K]
3 years ago
10

what type of boundary occurs where two plates move together causing one plae to descend in the mantle beneath the other plate

Physics
2 answers:
kvv77 [185]3 years ago
8 0

convergent boundary is the correct answer

irinina [24]3 years ago
7 0
This is called a subduction zone. The two plates converge, but one is forced under the other.
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Who is Janna Levin:
viva [34]

Answer:

Janna levin is a cosmologist and professor at physics.

She is an american by race

She was the presenter of Nova feature Black hole Apocalypse and has writtenany science non-fiction books

www.jannalevin.com is her own page where u get her correct info and bio

8 0
3 years ago
A 1.15-kg mass oscillates according to the equation where x is in meters and in seconds. Determine (a) the amplitude, (b) the fr
ANEK [815]

The complete question is;

A 1.15-kg mass oscillates according to the equation x = 0.650 cos(8.40t) where x is in meters and t in seconds. Determine (a) the amplitude, (b) the frequency, (c) the total energy, and (d) the kinetic energy and potential energy when x = 0.360 m.

Answer:

A) Amplitude; A = 0.650 m

B) Frequency; f = 1.337 Hz

C) total energy = 17.142 J

D) Kinetic energy = 11.884 J

Potential Energy = 5.258 J

Explanation:

We are given;

Mass;m = 1.15 kg

Equation; x = 0.650 cos (8.40t)

(a) The standard form of a wave function is in the form y(x,t) = Asin(kx−ωt+ϕ)

So, comparing terms in our equation in the question to this, the amplitude is;

A = 0.650 m

(b) we know that formula for frequency is;

f = ω/2π

Again, comparing terms in the standard equation and our question, we can see that ω = 8.4

Thus;

f = 8.4/(2π)

f = 1.337 Hz

(c) Formula for the total energy is given by;

E = m•ω²•A²/2

Plugging in the relevant values, we have;

E = (1.15)(8.40)²(0.650)²/2

E = 17.142 J

(d) we want to find the kinetic energy and potential energy when x = 0.360 m.

The formula for kinetic energy in this case is given by;

K = (1/2)•m•ω²•(A² - x²)  

Thus;

K = (1/2) × (1.15) × (8.40)² × ((0.650)² - (0.360)²)

K = 11.884 J

Also, the formula for the potential energy in this case is given by;

U = (1/2)•m•ω²•x²              

Thus;

U = (1/2) × (1.15) × (8.40)² × (0.360)²

U = 5.258 J

3 0
3 years ago
Part F What are some potential applications of a material that make things invisible?​
Sveta_85 [38]

Answer:

Its ahhshshgd

Explanation:

Very welcome mate.

8 0
2 years ago
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Sedimentary rocks are the only rocks that can potentially containa.fossils.c.fractures.b.minerals.d.faults. Please select the be
Elena-2011 [213]
Hello <span>Mr1guy24 </span>


Question: <span>Sedimentary rocks are the only rocks that can potentially contain?
</span>
Answer: Fossils (A)

Reason: This is what makes sedimentary rocks unique 

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8 0
3 years ago
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A section of a parallel-plate air waveguide with a plate separation of 7.11 mm is constructed to be used at 15 GHz as an evanesc
adell [148]

Answer:

the required minimum length of the attenuator is 3.71 cm

Explanation:

Given the data in the question;

we know that;

f_{c_1 = c / 2a

where f is frequency, c is the speed of light in air and a is the plate separation distance.

we know that speed of light c = 3 × 10⁸ m/s = 3 × 10¹⁰ cm/s

plate separation distance a = 7.11 mm = 0.0711 cm

so we substitute

f_{c_1 = 3 × 10¹⁰ / 2( 0.0711  )

f_{c_1 = 3 × 10¹⁰ cm/s / 0.1422  cm

f_{c_1 =  21.1 GHz which is larger than 15 GHz { TEM mode is only propagated along the wavelength }

Now, we determine the minimum wavelength required

Each non propagating mode is attenuated by at least 100 dB at 15 GHz

so

Attenuation constant TE₁ and TM₁ expression is;

∝₁ = 2πf/c × √( (f_{c_1 / f)² - 1 )

so we substitute

∝₁ = ((2π × 15)/3 × 10⁸ m/s) × √( (21.1 / 15)² - 1 )

∝₁ = 3.1079 × 10⁻⁷

∝₁ = 310.79 np/m

Now, To find the minimum wavelength, lets consider the design constraint;

20log₁₀e^{-\alpha _1l_{min = -100dB

we substitute

20log₁₀e^{-(310.7np/m)l_{min = -100dB

l_{min = 3.71 cm

Therefore, the required minimum length of the attenuator is 3.71 cm

6 0
3 years ago
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