Is a electron a positively charged, negatively charged, or a neutral particle

A) A wire made from iron with a cross-section of diameter 0.800 mm carries a current of 14.0 A. Calculate the "areal current den

Veseljchak [2.6K]

Answer:

A) ρ= $1.74x10^{26}$

B) μ= $1.68x10^{29}\frac{electron}{m^3}$

C) v= $1.03x10^{-3}\frac{m}{s}$

D)e= $8.99x10^-9$

Explanation:

A)

The magnetic field can be find knowing the current is the charge per second

β= $\frac{14eC*s}{1.6x10^{-19}C}\\$

β= 8.75x10^{19}e*s

Electron density

ρ= $\frac{8.75x10^{19}}{\pi*0.400x10^{-3}m} = 1.74x10^{26}$

B)

μ= $\frac{7.86}{56.2}\frac{g}{cm^3} \frac{mol}{g}*6.022x10^{23}\frac{molecules}{mol} *2 \frac{electron}{molecule}$

μ $=1.68x10^{23} \frac{electron}{cm^3}= 1.68x10^{29} \frac{electron}{m^3}$

C)

The drift speed using last information found

$v= \frac{J}{n*q} \\v= \frac{14A}{\pi*((0.40x10^-3)^2)*1.68x10^29*1.60x10^-19)} = 1.03x10^-3(\frac{m}{s} )$

D)

To compared the random thermal motion and the current's drift speed

$e=\frac{1.03x10^-3}{1.15x10^5} = 8.99x10^-9$

8 0

3 years ago

Wan and Nurul sat on a see-saw. The see-saw is imbalanced because Wan’s mass is 45 kg while Nurul’s mass is only 30 kg. Suggest

Alex Ar [27]

We have that the see-saw will be balance if a weight of $x=15kg$ is added to Nural's side of the see-saw

$x=15kg$

From the Question we are told that

Wan’s mass is $M_w=45 kg$

Nurul’s mass is $M_n= 30 kg.$

Generally

The Will be balance when the weight on both sides of the see-saw are equal

$M_w=M_n+x$

$45=30+x$

$x=45-30$

$x=15kg$

In conclusion

The see-saw will be balance if a weight of $x=15kg$ is added to Nural's side of the see-saw

$x=15kg$

For more information on this visit

brainly.com/question/22255610

5 0

3 years ago

Kirk wants to know if someone his height will go faster on a bike with a larger frame. What should he do next in order to follow

bearhunter [10]

Answer:

The answer is A but I could be wrong

8 0

3 years ago

Read 2 more answers

If 6000 J of heat is added to 200 gm of water at 25° C. What will be its final<br>temperature?

Debora [2.8K]

Answer:

T₂ = 305.17 K

Explanation:

Given that,

Heat, Q = 6000 J

Mass, m = 200 gram

Initial temperature, T₁ = 25° C

We need to find its final temperature. Let it is T₂.

We know that,

$Q=mc\Delta T$

Where

c is the specific heat of water, c = 4.18 J/g°C

So,

$6000=200\times 4.18\times (T_2-298)\\\\\dfrac{6000}{200\times 4.18}=(T_2-298)\\\\7.17=(T_2-298)\\\\7.17+298=T_2\\\\T_2=305.17\ K$

So, the final temperature is equal to 305.17 K.

3 0

3 years ago

Usually, when the temperture is increased l, what will happen to the ratevof dissolving?

pochemuha

<span>When temperature is increased, the rate of dissolving increases. The kinetic energy of the molecules of the solute and solvent molecules is high thereby increasing their contact. An example is mixing powdered sugar to the water. When you add water to the sugar, the dissolving process is slow. However, when you increase the temperature of the water by boiling it, the sugar dissolves immediately. </span>

3 0

3 years ago