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2 years ago

I need help on a english book report

Physics

1 answer:

lianna [129]2 years ago

7 0

Answer:

I can help, but

Explanation:

Where is the question?

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A small but measurable current of 3.8 × 10-10 A exists in a copper wire whose diameter is 2.5 mm. The number of charge carriers

Karolina [17]

Answer:

a) 4.9*10^-6

b) 5.71*10^-15

Explanation:

Given

current, I = 3.8*10^-10A

Diameter, D = 2.5mm

n = 8.49*10^28

The equation for current density and speed drift is

J = I/A = (ne) Vd

A = πD²/4

A = π*0.0025²/4

A = π*6.25*10^-6/4

A = 4.9*10^-6

Now,

J = I/A

J = 3.8*10^-10/4.9*10^-6

J = 7.76*10^-5

Electron drift speed is

J = (ne) Vd

Vd = J/(ne)

Vd = 7.76*10^-5/(8.49*10^28)*(1.60*10^-19)

Vd = 7.76*10^-5/1.3584*10^10

Vd = 5.71*10^-15

Therefore, the current density and speed drift are 4.9*10^-6

And 5.71*10^-15 respectively

3 0

3 years ago

Calculate the molecular weight of Aluminium hydroxide

Vedmedyk [2.9K]

Al(OH)3 = 26.98 + [(16×3) + (1.01×3)] = 26.98 + 51.03 = 78.01 and the unit will be g/mol

5 0

3 years ago

For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol. Assuming one free electron per copper atom, what is the drift velocity of electro

viktelen [127]

Answer:

$V_d$ = 1.75 × 10⁻⁴ m/s

Explanation:

Given:

Density of copper, ρ = 8.93 g/cm³

mass, M = 63.5 g/mol

Radius of wire = 0.625 mm

Current, I = 3A

Area of the wire, $A = \frac{\pi d^2}{4}$ = $A = \frac{\pi 0.625^2}{4}$

Now,

The current density, J is given as

$J=\frac{I}{A}=\frac{3}{ \frac{\pi 0.625^2}{4}}$ = 2444619.925 A/mm²

now, the electron density, $n = \frac{\rho}{M}N_A$

where,

$N_A$ =Avogadro's Number

$n = \frac{8.93}{63.5}(6.2\times 10^{23})=8.719\times 10^{28}\ electrons/m^3$

Now,

the drift velocity, $V_d$

$V_d=\frac{J}{ne}$

where,

e = charge on electron = 1.6 × 10⁻¹⁹ C

thus,

$V_d=\frac{2444619.925}{8.719\times 10^{28}\times (1.6\times 10^{-19})e}$ = 1.75 × 10⁻⁴ m/s

4 0

3 years ago

Read 2 more answers

How should the student change the circuit to give negative values for current and

dmitriy555 [2]

Answer:

Flip the cell.

Explanation:

This reverses direction of energy transfer.

Alternatively, flip ammeters and voltmeters to give negative readings.

what do penguins eat for lunch?

Ice-burgers!

3 0

3 years ago

3. A concrete highway is built of slabs 12 m long (20o C). How wide should the expansion cracks between the slabs be (at 20o C)

lesantik [10]

Answer:

$\Delta x=2.304\times 10^{-2}\ m=2.304\ cm$ is the minimum gap between the slabs

Explanation:

Given:

length of the concrete highway, $l=12\ m$

coefficient of thermal expansion, $\alpha=12\times 10^{-6}\ ^{\circ}C^{-1}$

range of temperature variation, $(-30^{\circ}C\ to\ 50^{\circ}C) \Rightarrow \Delta T=80^{\circ}C$

<u>Now form the equation of thermal expansion:</u>

$\Delta l=l\times \alpha\times \Delta T$

$\Delta l=12\times (12\times 10^{-6})\times 80$

$\Delta l=1.152\times 10^{-2}\ m$

Since each slab of the highway expands by the above length so the minimum gap between the slabs to prevent buckling:

$\Delta x=2\times \Delta l$

$\Delta x=2\times( 1.152\times 10^{-2})$

$\Delta x=2.304\times 10^{-2}\ m=2.304\ cm$ is the minimum gap between the slabs

4 0

3 years ago