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2 years ago

New seafloor material is formed at places in Earth’s crust where –

Physics

2 answers:

mr_godi [17]2 years ago

4 0

Answer:

D, I think.

Explanation:

I had a quiz in Plate Tectonics and there was 2 questions that are related to this, but not the exact question.

Which material rises from cracks in oceanic crust

-molten rock

Which is the first step in the seafloor spreading process?

-a crack forms in oceanic crust.

those are all right btw, so you can decide if the answer I told you is right or not.

nlexa [21]2 years ago

3 0

Answer:

Explanation:

Give the other person brainliest he has a good explanation.

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Look around the room you are in. Name

kozerog [31]

Answer:

A wire,rubber

Explanation:

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3 years ago

Read 2 more answers

Who performed classic experiments that supported the semiconservative model of dna replication?

Salsk061 [2.6K]

Meselson and Stahl

Explanation:

The classic experiment that supported the semiconservative model of dna replication was performed by Matthew Meselson and Franklin W. Stahl. In this model, the two strands of DNA unwind from each other, and each acts as a template for synthesis of a new, complementary strand. This results in two DNA molecules with one original strand and one new strand. They used E. coli bacteria as a model system.

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3 years ago

Two ships of equal mass are 110 m apart. What is the acceleration of either ship due to the gravitational attraction of the othe

dusya [7]

Answer:

Acceleration of the ship, $a=2.14\times 10^{-7}\ m/s^2$

Explanation:

It is given that,

Mass of both ships, $m=39000\ metric\ tons=39\times 10^6\ kg$

Distance between two ships, d = 110 m

The gravitational force between two ships is given by :

$F=G\dfrac{m^2}{d^2}$

$F=6.67\times 10^{-11}\ Nm^2/kg^2\times \dfrac{(39\times 10^6\ kg)^2}{(110\ m)^2}$

F = 8.38 N

Let a is the acceleration. Now, using second law of motion as :

$a=\dfrac{F}{m}$

$a=\dfrac{8.38\ N}{39\times 10^6\ kg}$

$a=2.14\times 10^{-7}\ m/s^2$

So, the acceleration of either ship due to the gravitational attraction of the other is $2.14\times 10^{-7}\ m/s^2$ . Hence, this is the required solution.