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2 years ago

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this is an area where two tectonic plates move towards each other with one going under the other plate?

1 answer:

Doss [256]2 years ago

6 0

Answer:Convergent boundaries are areas where plates move toward each other and collide. These are also known as compressional or destructive boundaries. Subduction zones occur where an oceanic plate meets a continental plate and is pushed underneath it. Subduction zones are marked by oceanic trenches.

pLS THANK ME AND VOTE ME BRAINLIEST ME

Explanation:

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In a transpiration experiment, the air bubble has an initial volume of 4.33 mL, and an initial pressure of 101.9 kPa. What will

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When a piece of copper is taken to the moon , a change will be observed in it's ?

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2 years ago

Three blocks are placed in contact on a horizontal frictionless surface. A constant force of magnitude F is applied to the box o

Lina20 [59]

Answer:

A) M

Explanation:

The three blocks are set in series on a horizontal frictionless surface, whose mutual contact accelerates all system to the same value due to internal forces as response to external force exerted on the box of mass M (Newton's Third Law). Let be F the external force, and F' and F'' the internal forces between boxes of masses M and 2M, as well as between boxes of masses 2M and 3M. The equations of equilibrium of each box are described below:

Box with mass M

$\Sigma F = F - F' = M\cdot a$

Box with mass 2M

$\Sigma F = F' - F'' = 2\cdot M \cdot a$

Box with mass 3M

$\Sigma F = F'' = 3\cdot M \cdot a$

On the third equation, acceleration can be modelled in terms of F'':

$a = \frac{F''}{3\cdot M}$

An expression for F' can be deducted from the second equation by replacing F'' and clearing the respective variable.

$F' = 2\cdot M \cdot a + F''$

$F' = 2\cdot M \cdot \left(\frac{F''}{3\cdot M} \right) + F''$

$F' = \frac{5}{3}\cdot F''$

Finally, F'' can be calculated in terms of the external force by replacing F' on the first equation:

$F - \frac{5}{3}\cdot F'' = M \cdot \left(\frac{F''}{3\cdot M} \right)$

$F = \frac{5}{3} \cdot F'' + \frac{1}{3}\cdot F''$

$F = 2\cdot F''$

$F'' = \frac{1}{2}\cdot F$

Afterwards, F' as function of the external force can be obtained by direct substitution:

$F' = \frac{5}{6}\cdot F$

The net forces of each block are now calculated:

Box with mass M

$M\cdot a = F - \frac{5}{6}\cdot F$

$M\cdot a = \frac{1}{6}\cdot F$

Box with mass 2M

$2\cdot M\cdot a = \frac{5}{6}\cdot F - \frac{1}{2}\cdot F$

$2\cdot M \cdot a = \frac{1}{3}\cdot F$

Box with mass 3M

$3\cdot M \cdot a = \frac{1}{2}\cdot F$

As a conclusion, the box with mass M experiments the smallest net force acting on it, which corresponds with answer A.

8 0

3 years ago

If the car’s speed decreases at a constant rate from 60 mi/hmi/h to 50 mi/hmi/h in 3.0 ss, what is the magnitude of its accelera

Kitty [74]

Answer:

The magnitude of the acceleration is 1.2 × 10⁴ mi/h²

Explanation:

Hi there!

The acceleration is defined as the change in velocity in a time:

a = Δv / Δt

Where:

a = acceleration.

Δv = change in velocity = final velocity - initial velocity.

Δt = elapsed time.

In this case:

Initial velocity = 60 mi/h

final velocity = 50 mi/h

elapsed time = 3.0 s

Let´s convert the time unit into h:

3.0 s · 1 h /3600 s = 1/1200 h

Now, let´s calculate the acceleration:

a = Δv / Δt

a = (50 mi/h - 60 mi/h) / 1/1200 h

a = -1.2 × 10⁴ mi/h²

The magnitude of the acceleration is 1.2 × 10⁴ mi/h²

7 0

3 years ago