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3 years ago

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If an atom has twelve electrons, how many electrons are in the n=1 energy level? how many electrons are in the n=2 energy level?

how many electrons are in the n=3 energy level?

1 answer:

Sav [38]3 years ago

3 0

The shells correspond with the principal quantum numbers (n<span> = </span>1<span>, </span>2<span>, </span>3<span>, 4 ...) Each shell can contain only a fixed number </span>of electrons. The first shell can hold up to two electrons<span>, the second shell can hold up to eight </span>electrons<span>, the third shell can hold up to 18 and so on.</span>

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What is the ocean floor made of?

Licemer1 [7]

Answer:

Oceanic crust is about 6 km (4 miles) thick. It is composed of several layers, not including the overlying sediment. The topmost layer, about 500 metres (1,650 feet) thick, includes lavas made of basalt (that is, rock material consisting largely of plagioclase [feldspar] and pyroxene).

Explanation:

it was on google hope this helps

3 0

4 years ago

A person throws a stone from the corner edge of a building. The stone's initial velocity is 28.0 m/s directed at 43.0° above the

Naya [18.7K]

The stone's acceleration, velocity, and position vectors at time $t$ are

$\mathbf a(t)=-g\,\mathbf j$

$\mathbf v(t)=v_{i,x}\,\mathbf i+\left(v_{i,y}-gt\right)\,\mathbf j$

$\mathbf r(t)=v_{i,x}t\,\mathbf i+\left(y_i+v_{i,y}t-\dfrac g2t^2\right)\,\mathbf j$

where

$g=9.80\dfrac{\rm m}{\mathrm s^2}$

$v_{i,x}=\left(28.0\dfrac{\rm m}{\rm s}\right)\cos43.0^\circ\approx20.478\dfrac{\rm m}{\rm s}$

$v_{i,y}=\left(28.0\dfrac{\rm m}{\rm s}\right)\sin43.0^\circ\approx19.096\dfrac{\rm m}{\rm s}$

and $y_i$ is the height of the building and initial height of the rock.

(a) After 6.1 s, the stone has a height of 5 m. Set the vertical component $(\mathbf j)$ of the position vector to 5 m and solve for $y_i$ :

$5\,\mathrm m=y_i+\left(19.096\dfrac{\rm m}{\rm s}\right)(6.1\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.1\,\mathrm s)^2$

$\implies\boxed{y_i\approx70.8\,\mathrm m}$

(b) Evaluate the horizontal component $(\mathbf i)$ of the position vector when $t=6.1\,\mathrm s$ :

$\left(20.478\dfrac{\rm m}{\rm s}\right)(6.1\,\mathrm s)\approx\boxed{124.92\,\mathrm m}$

(c) The rock's velocity vector has a constant horizontal component, so that

$v_{f,x}=v_{i,x}\approx20.478\dfrac{\rm m}{\rm s}$

where $v_{f,x}$

For the vertical component, recall the formula,

${v_{f,y}}^2-{v_{i,y}}^2=2a\Delta y$

where $v_{i,y}$ and $v_{f,y}$ are the initial and final velocities, $a$ is the acceleration, and $\Delta y$ is the change in height.

When the rock hits the ground, it will have height $y_f=0$ . It's thrown from a height of $y_i$ , so $\Delta y=-y_i$ . The rock is effectively in freefall, so $a=-g$ . Solve for $v_{f,y}$ :

${v_{f,y}}^2-\left(19.096\dfrac{\rm m}{\rm s}\right)^2=2(-g)(-124.92\,\mathrm m)$

$\implies v_{f,y}\approx-53.039\dfrac{\rm m}{\rm s}$

(where we took the negative square root because we know that $v_{f,y}$ points in the downward direction)

So at the moment the rock hits the ground, its velocity vector is

$\mathbf v_f=\left(20.478\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(-53.039\dfrac{\rm m}{\rm s}\right)\,\mathbf j$

which has a magnitude of

$\|\mathbf v_f\|=\sqrt{\left(20.478\dfrac{\rm m}{\rm s}\right)^2+\left(-53.039\dfrac{\rm m}{\rm s}\right)^2}\approx\boxed{56.855\dfrac{\rm m}{\rm s}}$

(d) The acceleration vector stays constant throughout, so

$\mathbf a(t)=\boxed{-g\,\mathbf j}$

4 0

3 years ago

Plzz help me with this

Ilya [14]

Answer: d

Explanation:

4 0

3 years ago

Which of the following is least likely to result from seafloor spreading

Assoli18 [71]

Answer:

An reversal in the magnetic fields of the north and south pole. This would be the most logical option for me...correct me if I'm wrong.

Explanation:

New seafloor is formed when magma is forced upward toward the surface at a mid-ocean.

7 0

3 years ago

A particle covers 10m distance in 2sec. if final velocity is 8m/s, find initial velocity

klasskru [66]

Explanation:

It is given that a particle covers 10m in first 5s and 10m in next 3s. so using the equation of motion

Case I

s=ut+

2

1

at

2

10=5u+

2

1

a(5)

2

20=10u+25a

4=2u+5a..............(1)

Case 2

In next 3s the particle covers more 10m distance. So

20=8u+

2

1

a(8)

2

5=2u+8a.........(2)

On solving equation (1) and (2)

4=2u+5a

5=2u+8a

a=

3

1

m/s

2

Put the value of a in equation (1)

u=

6

7

m/s

Now to find distance in next 10 s. total time will be 10s

s=

6

7

×10+

2

1

×

3

1

×(10)

2

s=28.33m

Distance travelled in next 2 sec

s=28.33−20=8.33m

4 0

3 years ago

Read 2 more answers