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3 years ago

How can gravity's role in tectonic plate motion be described

2 answers:

6 0

In ridge push, the mantle wells upward because of the convection and elevates the edges of spreading oceanic plates. Because these plates are higher at the spreading center, they are forced downhill due to <em>gravity</em> and eventually flatten out to the ocean floor.

Leya [2.2K]3 years ago

5 0

One of the main driving force for tectonic plate motion is gravity.

The subduction is the process where under gravity the oceanic lithosphere dives beneath another plate and thats how one plate meets another plate.This sink of oceanic lithosphere is responsible for dragging of rest of the tectonic plates. This is how the plates move.

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Explain the difference between a base unit and a derived unit

Scrat [10]

<span>A SI base unit is the single and direct measurement unit for a physical entity. E.g. For mass it is kg, for time it is sec, A derived unit is determined by a physical equation with the base units, like for velocity equals distance divided by time, in SI units: unit of v (velocity)= m (meter) divided by sec (seconds).

Hope this helps!</span>

3 0

3 years ago

Read 2 more answers

Since the half-life of 235U (7. 13 x 108 years) is less than that of 238U (4.51 x 109 years), the isotopic abundance of 235U has

Ymorist [56]

Answer:

$\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

Explanation:

Given that:

The Half-life of $^{235}U$ = $7.13 \times 10^8 \ years$ is less than that of $^{238} U = 4.51 \times 10^9 \ years$

Although we are not given any value about the present weight of $^{235}U$ .

So, consider the present weight in the percentage of $^{235}U$ to be y%

Then, the time elapsed to get the present weight of $^{235}U$ = $t_1$

Therefore;

$N_1 = N_o e^{-\lambda \ t_1}$

here;

$N_1$ = Number of radioactive atoms relating to the weight of y of $^{235}U$

Thus:

$In( \dfrac{N_1}{N_o}) = - \lambda t_1$

$In( \dfrac{N_o}{N_1}) = \lambda t_1$ --- (1)

However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of $^{235}U$ to be = $t_2$

Then:

$In( \dfrac{N_o}{N_2}) = \lambda t_2$ ---- (2)

here;

$N_2$ = Number of radioactive atoms of $^{235}U$ relating to 3.0 a/o weight

Now, equating equation (1) and (2) together, we have:

$In( \dfrac{N_o}{N_1}) -In( \dfrac{N_o}{N_2}) = \lambda( t_1-t_2)$

replacing the half-life of $^{235}U$ = $7.13 \times 10^8 \ years$

$In( \dfrac{N_2}{N_1}) = \dfrac{In 2}{7.13 \times 10^9}( t_1-t_2)$ ( since $\lambda = \dfrac{In 2}{t_{1/2}}$ )

∴

$\mathtt{In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2}= t_1-t_2}$

The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o

Thus, The time elapsed is $\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

8 0

3 years ago

An oxygen atom has a mass of 2.66 x 10^-23 g and a glass of water has a mass of 0.050kg. Use this information to answer the ques

fomenos

<h2>ANSWER OF EACH PART ARE GIVEN BELOW</h2>

Explanation:

We know, each mole contains $N_A=$ $6.023 \times 10^{23}$ atoms.

It is given that mass of one oxygen atom is m= $2.66\times 10^{-23}\ g$ .

Therefore, mass of one mole of oxygen, $M=m\times N_A$ .

Putting value of n and $N_A$ ,

$M=2.66\times 10^{-23}\times 6.023\times 10^{23} \ gm\\M=16.0\ gm$

Given,

Mass of water in glass=0.050 kg = 50 gm.

From above part mass of one mole of oxygen atoms = 16.0 gm.

Therefore, number of mole of oxygen equivalent to 50 gm oxygen $=\dfrac{50}{16}=3.1 \ moles.$

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3 years ago

When you multiply measurements, the answer should only have the same number of significant figures as the measurement with the _

lisov135 [29]

Answer: fewest

Explanation:

3 0

2 years ago

a 1.00litre vessel contains 0.215 mole of nitrogen gas and 0.0118 mole of hydrogen gas at 25°C. determine the partial pressure o

stepladder [879]

Answer:

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8 0

2 years ago