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Norma-Jean [14]
3 years ago
10

Suppose you have a large boulder in your yard that you’d like to move. How could you break it into smaller pieces without using

a sledgehammer?
Physics
2 answers:
Scrat [10]3 years ago
8 0

Is the question that is trying to be answered here how to move the boulder, or how to break up a boulder? If the purpose of the question is how to move the boulder from A to B, then you roll it.

laiz [17]3 years ago
8 0

-- Build a long, skinny campfire all around the rock, as close as possible to the rock, so the hot coals are right around the underside of the rock, and the flames lick up and around it.

-- Let it burn there for a few hours, so that a lot of the outside of the rock gets really hot.

-- Throw a big bunch of cold water on top of the rock ... enough so that it runs down the whole outside of the rock.

The thermal stresses that suddenly propagate through the rock, on the surface and for some distance into it, crack a layer of some thickness into several pieces, and the pieces fall off onto the ground.

If the cracking and crumbling doesn't extend far enough in, then just pick up the pieces of the layer that DID crumble off, and then go through the  whole same process again, with the inside smaller rock that remains.

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which observation about the Mid-Atlantic Ridge region provides the best evidence of the seafloor has been spreading over for mil
Kitty [74]

Magnetic stripes on the sea floor

Explanation:

The observation of magnetic stripes patterns at the region of the Mid-Atlantic Ridge galvanized the evidence that sea floor spreading has been on for several millions of years.

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Sea floor spreading brainly.com/question/9912731

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3 years ago
An astronaut is on a 100-m lifeline outside a spaceship, circling the ship with an angular speed of
nataly862011 [7]

Answer:

D = 72.68 m

Explanation:

given,

R = 100 m

angular speed = 0.1 rad/s

distance she can be pulled before the centripetal acceleration reaches 5g = 49 m/s².

using conservation of Angular momentum

I_i\omega_i= I_f\omega_f

mr_i^2\omega_i=m r_f^2\omega_f

\omega_f = \dfrac{r_i^2}{r_f^2}\times \omega_i

\omega_f = \dfrac{r_i^2}{r_f^2}\times \omega_i

we know,

centripetal acceleration

a = \dfrac{v^2}{r}

v = r ω

a =\omega_f^2 r_f

a =(\dfrac{r_i^2}{r_f^2}\times \omega_i)^2 r_f

a =\dfrac{r_i^4\times \omega_i^2}{r_f^3}

r_f^3=\dfrac{100^4\times 0.1^2}{5\times 9.8}

r_f^3=20408.1632

r_f = 27.32\ m

distance she has reached inward is equal to

D = 100 - 27.23

D = 72.68 m

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