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boyakko [2]
3 years ago
13

Enter the expression

Physics
1 answer:
vlabodo [156]3 years ago
6 0
I see the word "when..." kind of fading out at the end of the first line.
Whatever comes after it may be important.

If you're just supposed to copy the expression into the box,
then the problem is that you left the 'e' out of it.

I'm guessing that you're supposed to enter whatever the expression becomes
when either  N₀ or  ' t ' has some special value that's in the first line.

Just taking a wild guess here . . . . .

If it's  "Enter the expression ..... , when t=0 ." ,
then the correct answer in the box is     N₀  .

But that's just a wild guess.  As I pointed out, you cut off
the picture in the middle of the word 'when', and I've got
a hunch that there's something important after it.
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Answer:

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Explanation:

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Answer:

  • zero

Explanation:

m_l     is the magnetic quantum number.

The only possible value for the magnetic quantum number for an electron in an s orbital is 0.

The first three quantun numbers are:

  • n: principal quantum number. It may have positive integer values: 1, 2, 3, 4,5, 6, 7, ...

  • l : Azimuthal or angular momentum quantum number. It may have integer values from 0 to n - 1.

       This quantum number is related to the type (or shape) of the orbital:

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         For f orbitals l=3

In this case, it is an s orbital, so we have l=0.

  • m_l , the third quantum number can have integer values  {from-l}   to    {+l}

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What’s the question? Is it true or false?
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Water flows straight down from an open faucet. The cross-sectional area of the faucet is 2.4 × 10-4m2 and the speed of the water
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To solve this problem it is necessary to apply the continuity equations in the fluid and the kinematic equation for the description of the displacement, velocity and acceleration.

By definition the movement of the Fluid under the terms of Speed, acceleration and displacement is,

v_2^2 = v_1^2 + 2gh

Where,

V_i = Velocity in each state

g= Gravity

h = Height

Our values are given as,

A_1 = 2.4*10^{-4} m^2

v_1 = 0.8 m/s

h = 0.11m

Replacing at the kinetic equation to find V_2 we have,

v_2 = \sqrt{v_1^2 + 2gh}

v_2 = \sqrt{(0.8 m/s)^2 + 2(9.80 m/s2)(0.11 m)}

v_2= 1.67 m/s

Applying the concepts of continuity,

A_1v_1 = A_2v_2

We need to find A_2 then,

A_2= \frac{A_1v_1 }{v_2}

So the cross sectional area of the water stream at a point 0.11 m below the faucet is

A_2= \frac{A_1v_1 }{v_2}

A_2= \frac{(2.4*10^{-4})(0.8)}{(1.67)}

A_2= 1.14*10^{-4} m2

Therefore the cross-sectional area of the water stream at a point 0.11 m below the faucet is 1.14*10^{-4} m2

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