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3 years ago

Subduction occurs as a result of:__________

Physics

1 answer:

lutik1710 [3]3 years ago

6 0

Answer:

a) Gravity pulling the older and denser lithosphere downward.

Explanation:

Earth's crust is made of plates which float on mantle . Plates are of two types

1 ) oceanic plates which are older and heavier 2 ) continental plates which are lighter . Continents are situated on continental plates. Due to relative motion , oceanic plate, because of its heavy weight, goes under lighter continental plate . This is called subduction.

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Were the continents once joined together as a supercontinent? Give 3 pieces of evidence to support Alfred Wegeners Theory of Con

sattari [20]

Yes! Fossils, The outlines of the continents and geological features .

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3 years ago

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1. David and Mat are training for the swimming team, swimming lengths of 25 metre pool. Mat swims four lengths in 125 seconds at

FromTheMoon [43]

Answer:

a) Mat swum 100 meters.

b)after 4 length , Mat's displacement is zero.

Explanation:

As given, Mat and David are training for the swimming team , and Length of 1 side of the pool is 25 meter.

As swimming pool will be a cube or a cuboid , so it has only 4 sides.

1 length = 23 meter

⇒4 length = 4×25 = 100 meter

So, Mat swam 100 meters.

As after swam 4 lengths of the pool, Mat reaches to its initial position i.e Mat come back to its starting point.

As Displacement is always measures from the starting point.

So, after 4 length , Mat's displacement is zero.

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4 years ago

You are standing next to a table and looking down on a record player sitting on the table. Take the spindle (axis of rotation) t

VLD [36.1K]

Answer:

a) the rotational speed of the clay is 3.45 rad/s

b) the value of A in the equation of motion is 0.15 m

c) the value of ϕi is 90° or π/2 rad.

Explanation:

Given that;

Revolution per minute rpm = 33( 1/3) = 100/3

The frequency f = 100 / 3(60) = 0.55 Hz

Rotational speed W = 2πf

we substitute

W = 2π × 0.55

W = 3.45 rad/s

Therefore, the rotational speed of the clay is 3.45 rad/s

given equation; y(t)=Asin(ωt+ϕi)

given that radius = 0.15 m

y(t)=(0.2)sin(ωt+ϕi)

Therefore, the value of A in the equation of motion is 0.15 m

since y(t) has the maximum value at t =0

so at t=0

y(0) = (0.15)sin(ω(0)+ϕi)

= 0.15sin(ϕi)

this will give maximum value when ϕi = 90°

y(0) = (0.15)sin(ω(0)+ϕi)

= 0.15sin(90°)

= 0.15

hence, the value of ϕi is 90° or π/2 rad.

7 0

3 years ago

A 54.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 126 m/s from the top of a cliff

Nikolay [14]

Answer:

a) The initial total mechanical energy of the projectile is 498556.296 joules.

b) The work done on the projectile by air friction is 125960.4 joules.

c) The speed of the projectile immediately before it hits the ground is approximately 82.475 meters per second.

Explanation:

a) The system Earth-projectile is represented by the Principle of Energy Conservation, the initial total mechanical energy ( $E$ ) of the project is equal to the sum of gravitational potential energy ( $U_{g}$ ) and translational kinetic energy ( $K$ ), all measured in joules:

$E = U_{g} + K$ (Eq. 1)

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

$E = m\cdot g\cdot y + \frac{1}{2}\cdot m\cdot v^{2}$ (Eq. 1b)

Where:

$m$ - Mass of the projectile, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$y$ - Initial height of the projectile above ground, measured in meters.

$v$ - Initial speed of the projectile, measured in meters per second.

If we know that $m = 54\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y = 132\,m$ and $v = 126\,\frac{m}{s}$ , the initial mechanical energy of the earth-projectile system is:

$E = (54\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (132\,m)+\frac{1}{2}\cdot (54\,kg)\cdot \left(126\,\frac{m}{s} \right)^{2}$

$E = 498556.296\,J$

The initial total mechanical energy of the projectile is 498556.296 joules.

b) According to this statement, air friction diminishes the total mechanical energy of the projectile by the Work-Energy Theorem. That is:

$W_{loss} = E_{o}-E_{1}$ (Eq. 2)

Where:

$E_{o}$ - Initial total mechanical energy, measured in joules.

$E_{1}$ - FInal total mechanical energy, measured in joules.

$W_{loss}$ - Work losses due to air friction, measured in joules.

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

$W_{loss} = E_{o}-K_{1}-U_{g,1}$

$W_{loss} = E_{o} -\frac{1}{2}\cdot m\cdot v_{1}^{2}-m\cdot g\cdot y_{1}$ (Eq. 2b)

Where:

$m$ - Mass of the projectile, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$y_{1}$ - Maximum height of the projectile above ground, measured in meters.

$v_{1}$ - Current speed of the projectile, measured in meters per second.

If we know that $E_{o} = 498556.296\,J$ , $m = 54\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y_{1} = 297\,m$ and $v_{1} = 89.3\,\frac{m}{s}$ , the work losses due to air friction are:

$W_{loss} = 498556.296\,J -\frac{1}{2}\cdot (54\,kg)\cdot \left(89.3\,\frac{m}{s} \right)^{2} -(54\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (297\,m)$

$W_{loss} = 125960.4\,J$

The work done on the projectile by air friction is 125960.4 joules.

c) From the Principle of Energy Conservation and Work-Energy Theorem, we construct the following model to calculate speed of the projectile before it hits the ground:

$E_{1} = U_{g,2}+K_{2}+1.5\cdot W_{loss}$ (Eq. 3)

$K_{2} = E_{1}-U_{g,2}-1.5\cdot W_{loss}$

Where:

$E_{1}$ - Total mechanical energy of the projectile at maximum height, measured in joules.

$U_{g,2}$ - Potential gravitational energy of the projectile, measured in joules.

$K_{2}$ - Kinetic energy of the projectile, measured in joules.

$W_{loss}$ - Work losses due to air friction during the upward movement, measured in joules.

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

$\frac{1}{2}\cdot m \cdot v_{2}^{2} = E_{1}-m\cdot g\cdot y_{2}-1.5\cdot W_{loss}$ (Eq. 3b)

$m\cdot v_{2}^{2} = 2\cdot E_{1}-2\cdot m \cdot g \cdot y_{2}-3\cdot W_{loss}$

$v_{2}^{2} = 2\cdot \frac{E_{1}}{m}-2\cdot g\cdot y_{2}-3\cdot \frac{W_{loss}}{m}$

$v_{2} = \sqrt{2\cdot \frac{E_{1}}{m}-2\cdot g\cdot y_{2}-3\cdot \frac{W_{loss}}{m} }$

If we know that $E_{1} = 372595.896\,J$ , $m = 54\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y_{2} =0\,m$ and $W_{loss} = 125960.4\,J$ , the final speed of the projectile is:

$v_{2} =\sqrt{2\cdot \left(\frac{372595.896\,J}{54\,kg}\right)-2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0\,m)-3\cdot \left(\frac{125960.4\,J}{54\,kg}\right) }$

$v_{2} \approx 82.475\,\frac{m}{s}$

The speed of the projectile immediately before it hits the ground is approximately 82.475 meters per second.

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3 years ago

A sound-producing object is moving away from an observer. The sound the observer hears will have a frequency that actually being

kozerog [31]

Lower than, this is due to the doplar effect

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3 years ago

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