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3 years ago

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A tectonic plate is part of the Earth’s lithospherehttp://brainly.com/question/add?entry=1642&task_content=A%20tectonic%20pl

ate%20is%20part%20of%20the%20Earth%E2%80%99s%20lithosphere

1 answer:

lord [1]3 years ago

3 0

Plate tectonics<span>is a </span>scientific theory<span> that describes the large-scale motion of </span>Earth<span>'s </span>lithosphere<span>. This theoretical model builds on the concept of </span>continental drift<span> which was developed during the first few decades of the 20th century. The </span>geoscientific<span> community accepted plate-tectonic theory after </span>seafloor spreading was validated in the late 1950s and early 1960s.<span>The lithosphere, which is the rigid outermost shell of a planet (the crust and upper mantle), is broken up into </span>tectonic plates<span>. </span>

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The cloud of interstellar dust and gas that forms a star is known as a

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The cloud of interstellar dust and gas that forms a star is known as a nebula. With an average surface temperature of about 737K, venus is the hottest planet. Hope this helps, and sorry your question didn't get answered in time.

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3 years ago

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A very flexible helium-filled balloon is released from the ground into the air at 20. ∘C. The initial volume of the balloon is 5

koban [17]

Answer:

V = 38.0 L

Explanation:

As we know that number of moles will remains conserved inside the balloon

so we will have

$moles = \frac{PV}{RT}$

here we have

$\frac{P_1V_1}{RT_1} = \frac{P_2V_2}{RT_2}$

now we have

$P_1 = 760 mm Hg$

$P_2 = 76 mm Hg$

$V_1 = 5.00 L$

$T_1 = 20^o C = 293 K$

$T_2 = -50^o C = 223 K$

$\frac{(760mm Hg)(5L)}{R(293)} = \frac{(76mm Hg)(V)}{R(223)}$

$V = 38.0 L$

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3 years ago

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A ball is fixed to the end of a string, which is attached to the ceiling at point P. As the drawing shows, the ball is projected

Ganezh [65]

Answer:

The ball's initial kinetic energy

The ball comes to a stop at B. At this point its initial kinetic energy is converted into potential energy

Explanation:

A ball is fixed to the end of a string, which is attached to the ceiling at point P. As the drawing shows, the ball is projected downward at A with the launch speed v0. Traveling on a circular path, the ball comes to a halt at point B. What enables the ball to reach point B, which is above point A? Ignore friction and air resistance.

From conservation of energy which states that energy can neither be created nor be destroyed, but can be transformed from one form to another.

Ki+Ui=Kf+Uf

Ki=initial kinetic energy

Ui=initial potential energy

Kf=final kinetic energy

Uf=final potential energy

we know that $\frac{1}{2} mu^{2} +mgha=\frac{1}{2} mv^{2} +mghb$

m=mass of the ball

ha=downward height a

hb=upward height b

u=initial velocity u

v=final velocity v, which is 0

g=acceleration due to gravity

v=0 at final velocity

1/2mu^2+mgha=0+1/2mv^2

ha=hb+Ki/mh

From the above equation, we can conclude that the ball's initial kinetic energy is responsible for making the ball reach point B.

Point B is higher than point A from the motion gained by the ball

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3 years ago

A 1-kilogram mass is attached to a spring whose constant is 14 N/m, and the entire system is then submerged in a liquid that imp

ch4aika [34]

Answer:

Part(a): The equation of motion is $\bf{x(t) = \dfrac{7}{5}~e^{-2t} - \dfrac{2}{5}~e^{-7t}}$ .

Part(b): The equation of motion is $\bf{x(t) = -e^{-2t} + \dfrac{11}{5}~e^{-7t}}$ .

Explanation:

If ' $m$ ' be the mass of the object, ' $k$ ' be the force constant and ' $\beta$ ' be the damping constant, then the equation of motion of the particle can be written as

$\dfrac{d^{2}x}{dt^{2}} + \dfrac{\beta}{m} \dfrac{dx}{dt} + \dfrac{k}{m}x= 0.........................................(I)$

Given $m$ = 1 Kg, $k$ = 14 $N~m^{-1}$ , $\beta$ = 9. Substituting these values in equation (I),

$\dfrac{d^{2}x}{dt^{2}} + 9~\dfrac{dx}{dt} + 14~x= 0$

Taking a trial solution $x(t) = e^{mt}$ , the auxiliary equation can be written as

$m^{2} + 9m + 14 = 0............................................................(II)$

and its solutions are $m_{1} = -2~and~m_{2} = -7$ , resulting the general solution

$x(t) = C_{1}~e^{-2t} + C_{2}~e^{-7t}....................................................................(III)$

The velocity at any instant of time of the mass is

$v(t) = -2C_{1}~e^{-2t} _7~C_{2}~e^{-7t}..............................................................(IV)$

Part(a):

Given $x(t=0) = 1 m,~and~v(t=0) = 0~m~s^{-1}$ . Substituting these values in equation (III) and (IV),

$&& 1 = C_{1} + C_{2}......................(V)\\&and,& 0 = -2C_{1} - 7C_{2}.......................(VI)$

Solving equations (V) and (VI), we have

$C_{1} = \dfrac{7}{5}~and~C_{2} = \dfrac{-2}{5}$

So the equation of motion is

$x(t) = \dfrac{7}{5}~e^{-2t} - \dfrac{2}{5}~e^{-7t}$

Part(b):

Given $x(t=0) = 1 m,~and~v(t=0) = - 12~m~s^{-1}$ . Substituting these values in equation (III) and (IV),

$&& 1 = C_{1} + C_{2}......................(VII)\\&and,& -12 = -2C_{1} -7C_{2}.......................(VIII)$

Solving equations (V) and (VI), we have

$C_{1} = -1~and~C_{2} = \dfrac{11}{5}$

So the equation of motion is

$x(t) = -e^{-2t} + \dfrac{11}{5}~e^{-7t}$

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3 years ago

Imagine no girls awnser this question ://

Nat2105 [25]

Answer:

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Explanation:

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3 years ago

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