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3 years ago

What does fissure and conduit have in common?

Physics

1 answer:

Flura [38]3 years ago

8 0

Answer:

although commons their deposits can dense

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A(n) 7.7-kg object is sliding across the ice at 2.34 m/s in the positive x direction. An internal explosion occurs, splitting th

-BARSIC- [3]

Answer:

Average acceleration on first part of the chunk is given as

$a_1 = 13.125 m/s^2$

Average acceleration on second part of the chunk is given as

$a_2 = -13.125 m/s^2$

Explanation:

By momentum conservation along x direction we will have

$mv_i = \frac{m}{2}v_1 + \frac{m}{2}v_2$

so we have

$v_1 + v_2 = 2v$

$v_1 + v_2 = 4.68$

also by energy conservation

$\frac{1}{2}(\frac{m}{2})v_1^2 + \frac{1}{2}(\frac{m}{2})v_2^2 - \frac{1}{2}mv^2 = 17 J$

$\frac{1}{4}m(v_1^2 + v_2^2) - \frac{1}{2}mv^2 = 17$

$(v_1^2 + v_2^2) - 2v^2 = \frac{4}{7.7}(17)$

$(4.68 - v_2)^2 + v_2^2 - 2v^2 = 8.83$

$21.9 + 2v_2^2 - 9.36 v_2 - 10.95 = 8.83$

$2v_2^2 - 9.36v_2 + 2.12 = 0$

by solving above equation we will have

$v_1 = 4.44 m/s$

$v_2 = 0.24 m/s$

Average acceleration on first part of the chunk is given as

$a_1 = \frac{4.44 - 2.34}{0.16}$

$a_1 = 13.125 m/s^2$

Average acceleration on second part of the chunk is given as

$a_2 = \frac{0.24 - 2.34}{0.16}$

$a_2 = -13.125 m/s^2$

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3 years ago

Janet jumps off a high diving platform with a horizontal velocity of 2.06 m/s and lands in the water 1.8 s later. How high is th

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Initial height of platform is 15.87m or 16m.

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3 years ago

How do icebergs form

pishuonlain [190]

Answer:

They break off large ice sheets found at the North and South poles.

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2 years ago

A non-_____ rock has interlocking grains with no specific pattern.

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A non <span>foliated </span>rock has interlocking grains with no specific pattern.

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3 years ago

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5.An ice skater pushes against a wall with a force of 59 N. Ignoring friction, if the ice skater has

natali 33 [55]

Answer:

<em>Answer: (A) 0.75 m/s^2</em>

Explanation:

The Second Newton's law states that an object acquires acceleration when an external unbalanced net force is applied to it.

That acceleration is proportional to the net force and inversely proportional to the mass of the object.

It can be expressed with the formula:

$\displaystyle a=\frac{F_n}{m}$

Where

Fn = Net force

m = mass

The ice skater pushes against a wall with a force of 59 N. The wall returns the force and the skater now has a net force of Fn=59 N that makes him accelerate. Being m=79 kg the mass of the skater, the acceleration is:

$\displaystyle a=\frac{59}{79}$

$a = 0.75\ m/s^2$

Answer: (A) 0.75 m/s^2

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3 years ago