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2 years ago

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Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.75 A out of the jun

ction. How many electrons per second move past a point in wire 3?

1 answer:

AlexFokin [52]2 years ago

3 0

Answer:

number of electrons = 2.18*10^18 e

Explanation:

In order to calculate the number of electrons that move trough the second wire, you take into account one of the Kirchoff's laws. All the current that goes inside the junction, has to go out the junction.

Then, if you assume that the current of the wire 1 and 3 go inside the junction, then, all this current have to go out trough the second junction:

$i_1+i_3=i_2$ (1)

i1 = 0.40 A

i2 = 0.75 A

you solve the equation i3 from the equation (1):

$i_3=i_2-i_1=0.75A-0.40A=0.35A$

Next, you take into account that 1A = 1C/s = 6.24*10^18

Then, you have:

$0.35A=0.35\frac{C}{s}=0.35*\frac{6.24*10^{18}e}{s}=2.18*10^{18}\frac{e}{s}$

The number of electrons that trough the wire 3 is 2.18*10^18 e/s

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Assume the radius of an atom, which can be represented as a hard sphere, is r = 1.95 Å. The atom is placed in a (a) simple cubic

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Answer:

(a) A = $3.90 \AA$

(b) $A = 4.50 \AA$

(c) $A = 5.51 \AA$

(d) $A = 9.02 \AA$

Solution:

As per the question:

Radius of atom, r = 1.95 $\AA = 1.95\times 10^{- 10} m$

Now,

(a) For a simple cubic lattice, lattice constant A:

A = 2r

A = $2\times 1.95 = 3.90 \AA$

(b) For body centered cubic lattice:

$A = \frac{4}{\sqrt{3}}r$

$A = \frac{4}{\sqrt{3}}\times 1.95 = 4.50 \AA$

(c) For face centered cubic lattice:

$A = 2{\sqrt{2}}r$

$A = 2{\sqrt{2}}\times 1.95 = 5.51 \AA$

(d) For diamond lattice:

$A = 2\times \frac{4}{\sqrt{3}}r$

$A = 2\times \frac{4}{\sqrt{3}}\times 1.95 = 9.02 \AA$

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1 year ago

Dane is holding an 8 kilogram box 2 metres above the ground. How much energy is in the box's gravitational potential energy stor

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156.8 Joules of energy is in the box's gravitational potential energy store

<u>Explanation</u>:

<em>Given:</em>

Mass of the box Dane is holding = 8 Kilograms

Height at which Dane is holding the box above the ground= 2 metres

<em>To Find:</em>

Gravitational potential energy in the box=?

<em>Solution:</em>

gravitational potential energy is the work done per mass on a object to move that object from one fixed location to to another location against gravity.Its unit is joules or J

Thus Gravitational potential energy is represented as,

$PE_g=mgh$

where

$PE_g$ is the gravitational potential energy

m is the mass

h is the height

g is the gravitational force( 9.8 $m/s^2$ )

Now substituting the given values,

$PE_g=8\times 9.8\times 2$

$PE_g=156.8 Joules$

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