Some straight wave fronts hit a barrier as shown here. Which of the positions shown (A, B,

An 18 gauge copper wire (the size usually used for lamp cords), with a diameter of 1.02 mm,1.02 mm, carries a constant current o

larisa86 [58]

Answer:

J = 2.044x10⁶ A/m²

v = 1.50x10⁻⁴ m/s

Explanation:

The current density (J) of the copper wire is giving by:

$J = \frac {I}{A}$

where I: electric current and A: cross-sectional area of the copper wire

The cross-sectional area of the copper wire can be calculated by:

$A = \frac {\pi d^{2}}{4} = \frac {\pi (1.02 \cdot 10^{-3} m)^{2}}{4} = 8.17 \cdot 10^{-07} m^{2}$

Substituting the calculated area in the equation (1) we have:

$J = \frac {1.67 A}{8.17 \cdot 10^{-7} m^{2}} = 2.044 \cdot 10^{6} \frac {A}{m^{2}}$

Hence, the current density is 2.044x10⁶ A/m².

To find the drift speed (v), we need to use the next equation:

$v = \frac {J}{n q}$

where n: the free-electron density, q: module of the charge of the electron

$v = \frac {2.044 \cdot 10^{6} \frac {A}{m^{2}}}{(8.5 \cdot 10^{28} {m^{-3}}) (1.6 \cdot 10^{-19} C)}$

$v = 1.50 \cdot 10^{-04} \frac {m}{s}$

So, the drift speed is 1.50x10⁻⁴ m/s.

Have a nice day!

4 0

4 years ago

Read 2 more answers

A spring-launched toy is propelled into the air. At first when the spring is compressed and the toy is placed on top. Before the

wlad13 [49]

In this problem, you are asked to find a vertical position of a ball when you are given its initial position on a spring. In both locations, the speed of the ball is zero.

If non-conservative forces are either known or small and if energy is converted from one form to another between the locations, then any time you relate speed and position of an object at two different points, conservation of energy is the most direct way to understand the problem.

In this case, you start out with stored energy in the compression of the spring and convert it to stored gravitational energy.

3 0

3 years ago

What is the meaning of Constant Acceleration?

lora16 [44]

Any change in speed or direction of motion is acceleration.

Constant acceleration can mean ...

-- speeding up at a constant rate . . . gaining the same amount
of speed each second.

-- slowing down at a constant rate . . . losing the same amount
of speed each second.

-- changing direction at a constant rate . . . for example, going
around a circular path at a constant speed.

4 0

3 years ago

A 0.0140 kg bullet traveling at 205 m/s east hits a motionless 1.80 kg block and bounces off it, retracing its original path wit

makvit [3.9K]

Answer:

Final velocity of the block = 2.40 m/s east.

Explanation:

Here momentum is conserved.

Initial momentum = Final momentum

Mass of bullet = 0.0140 kg

Consider east as positive.

Initial velocity of bullet = 205 m/s

Mass of Block = 1.8 kg

Initial velocity of block = 0 m/s

Initial momentum = 0.014 x 205 + 1.8 x 0 = 2.87 kg m/s

Final velocity of bullet = -103 m/s

We need to find final velocity of the block( u )

Final momentum = 0.014 x -103+ 1.8 x u = -1.442 + 1.8 u

We have

2.87 = -1.442 + 1.8 u

u = 2.40 m/s

Final velocity of the block = 2.40 m/s east.

7 0

4 years ago

You observe a spiral galaxy with a large central bulge and tightly wrapped arms. It would be classified a

miv72 [106K]

Answer:

Sa

Explanation:

Spiral Galaxies -

It is a disk shaped galaxies which have spiral structure , is refereed to as spiral galaxies .

According to Hubble , these galaxies are classified as Sa , Sb , Sc .

Where ,

Sa - have the structure , which is bulged from the central portion , along with a tightly wrapped spiral structure .

Sb - have a lesser bulge and the spiral is looser .

Sc - It has very weak bulge with the open spiral structure .

Hence , from the question ,

The given information is about the Sa .

6 0

4 years ago