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3 years ago

15

This is because we see evidence of plate movement in the fact that as we look at rocks farther and farther from the ridge, the r

ocks--
A)Have higher iron content

B)Are made up of smaller particles

C)Become thicker and thicker

D)Are consistently older

2 answers:

Arisa [49]3 years ago

8 0

Answer:c

Explanation:

Flauer [41]3 years ago

6 0

Answer:

C

Explanation:

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Name all of the steps on the Physical Activity Pyramid?

iogann1982 [59]

Answer:

The steps are:-

Moderate physical activity
Vigorous aerobics
Vigorous sport and recreation
Muscle fitness exercise
Flexibility exercise
Avoiding inactivity
Balancing energy

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3 years ago

in volleyball if you are a server what position on the court are you? center back #6, right back #1 or left back #5? help me ple

jek_recluse [69]

In the sport of volleyball a server is in the "back left position" for standard rules.

Hope I helped,
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3 years ago

a ball is thrown straight up it passes a 2.00 m high window 7.5 meters off the ground on its path up it takes 1.3 s to go past t

Svet_ta [14]

Let's say the velocity at the bottom of the window was "v."
s = v*t + ½at²
2 m = v * 1.3s - 4.9m/s² * (1.3s)² = v * 1.3s - 8.3 m
v = 10.3m / 1.3s = 7.9 m/s

Then the initial speed was
V = √(v² + 2as) = √(7.9m/s² + 2 * 9.8m/s² * 7.5m) = 14 m/s ◄ initial velocity
(after rounding to 2 digits from 14.5 m/s).

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3 years ago

Does the number of particles change as the substance changes its state?

Leviafan [203]

Particles stay the same unless there is a chemical change whether the matter is solid, liquid or gas. ... When substances change state there is no change in mass so if 100 g of ice is melted 100g of water are formed this will boil to form 100g of steam (this is called "conservation of mass").

4 0

3 years ago

Q. No. 9 A body falls freely from the top of a tower and during the last second of its fall, it falls through 25m. Find the heig

HACTEHA [7]

Answer:

45.6m

Explanation:

The equation for the position y of an object in free fall is:

$y=-\frac{1}{2} gt^2+v_0t+y_0$

With the given values in the question the equation has one unknown v₀:

$v_0=\frac{y-y_0}{t}+\frac{1}{2}gt$

Solving for t=1:

1) $v_0=y-y_0+\frac{g}{2}$

To find the hight of the tower you can use the concept of energy conservation:

The energy of the body 1 sec before it hits the ground:

2) $E=\frac{1}{2}m{v_0}^2+mgy_0$

If h is the height of the tower, the energy on top of the tower:

3) $E=mgh$

Combining equation 2 and 3 and solving for h:

4) $h=\frac{{v_0}^2}{2g}+y_0$

Combining equation 1 and 4:

$h=\frac{{(y-y_0+\frac{g}{2}})^2}{2g}+y_0$

4 0

3 years ago