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3 years ago

5

How many electrons move past a fixed reference point every t = 2.55 ps if the current is i = 7.3 μA ? Express your answer as an

integer.

1 answer:

iris [78.8K]3 years ago

4 0

Answer:

116.3 electrons

Explanation:

Data provided in the question:

Time, t = 2.55 ps = 2.55 × 10⁻¹² s

Current, i = 7.3 μA = 7.3 × 10⁻⁶ A

Now,

we know,

Charge, Q = it

thus,

Q = (7.3 × 10⁻⁶) × (2.55 × 10⁻¹²)

or

Q = 18.615 × 10⁻¹⁸ C

Also,

We know

Charge of 1 electron, q = 1.6 × 10⁻¹⁹ C

Therefore,

Number of electrons past a fixed point = Q ÷ q

= [ 18.615 × 10⁻¹⁸ ] ÷ [ 1.6 × 10⁻¹⁹ ]

= 116.3 electrons

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Carbon dioxide steadily flows into a constant pressure heater at 300 K and 100 kPa witha mass flow rate of 9.2 kg/s. Heat transf

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Answer:

Carbon dioxide temperature at exit is 317.69 K

Carbon dioxide flow rate at heater exit is 20.25 m³/s

Explanation:

Detailed steps are attached below.

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3 years ago

A particular NMOS device has parameters VT N = 0.6 V, L = 0.8µm, tox = 200 Å, and µn = 600 cm2 /V–s. A drain current of ID = 1.2

NeTakaya

Answer:

$W= 3.22 \mu m$

Explanation:

the transistor In saturation drain current region is given by:

$i_D}=K_a(V_{GS}-V_{IN})^2$

Making $K_a$ the subject of the formula; we have:

$K_a=\frac {i_D} {(V_{GS} - V_{IN})^2}$

where;

$i_D = 1.2m$

$V_{GS}= 3.0V$

$V_{TN} = 0.6 V$

$K_a=\frac {1.2m} {(3.0 - 0.6)^2}$

$K_a = 208.3 \mu A/V^2$

Also;

$k'_n}=\frac{\mu n (\frac{cm^2}{V-s} ) \epsilon _{ox}(\frac{F}{cm} ) }{t_{ox}(cm)}$

where:

$\mu n (\frac{cm^2}{V-s} ) = 600$

$\epsilon _{ox}=3.9*8.85*10^{-14}$

${t_{ox}(cm)=200*10^{-8}$

substituting our values; we have:

$k'_n}=\frac{(600)(3.988.85*10^{-14})}{(200*10^{-8})}$

$k'_n}=103.545 \mu A/V^2$

Finally, the width can be calculated by using the formula:

$W= \frac{2LK_n}{k'n}$

where;

L = $0.8 \mu m$

$W= \frac{2*0.8 \mu m *208.3 \mu}{103.545 \mu}$

$W= 3.22 \mu m$

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4 years ago

In contrast to the leading-trailing drum brake system, the duo-servo drum brake system will:

prohojiy [21]

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4 years ago

The symmetrical load below is connected to a three-phase network. A line current of 25A has been measured. The load resistors ha

boyakko [2]

Answer:

The line voltage of the three phase network is 346.41 V

Explanation:

Star Connected Load

Resistance, R₁ = R₂ = R₃ = 18 Ω

For a star connected load, the line current = the phase current, that is we have

$I_L = 25 \, A = I_{Ph}$

Whereby the the voltage across each resistance = $V_R$ is given by the relation;

$V_R$ = $I_{Ph}$ × R

Hence;

$V_{Ph}$ = $V_R$ = $I_{Ph}$ × R = 25 × 8 = 200 V

Therefore we have;

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4 years ago

A metal having a cubic structure has a density of , an atomic weight of , and a lattice parameter of Å One atom is associated wi

Veronika [31]

Answer:

Explanation:

Answer: The crystal structure of the metal is BCC

Explanation:

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Volume of unit cell= (a°)^3.

The lattice parameter here is a°.

Substitute (6.13 * 10^-8)cm for a°.

Volume of unit cell = (6.13 * 10^-8)^3 = 2.3034 * 10^-22 cm^3/cell.

To determine the crystal structure we use

Density (p) = {(Number of atoms per cell) (Atomic mass)} / {(volume of unit cell)(Avogrado constant)}.

Substitute 1.892g/cm^3 for p (6.02*10^23) atoms/mol for Avogrado constant 1.3921g/mol.

For atomic mass and (2.3034 * 10^-22) cm^3/cell for unit cell.

1.892g/cm^3 = {(Number of atoms per cell) (1.3291g/mol)} / {(2.3034 * 10^-22) (6.02 * 10^23 atoms/mol)}.

Changing the subject of formula we have :

Number of atoms per cell = {(2.3034 * 10^-22) * (6.02 * 10^23) * 1.892} / 132.91

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Note: p = density

a° = a subscript o

4 0

3 years ago