1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Svetach [21]
3 years ago
14

The charge of an electron is -1.60x10-19 C. A current of 1 A flows in a wire carried by electrons. How many electrons pass throu

gh a cross section of the wire each second?
Physics
1 answer:
faltersainse [42]3 years ago
4 0

Answer: 6.241\times 10^{18} electrons pass through a cross section of the wire each second.

Explanation:

According to mole concept:

1 mole of an atom contains 6.022\times 10^{23} number of particles.

Given : Charge on 1 electron = 1.6\times 10^{-19}C

Charge on 1 mole of electrons = 1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C

To calculate the charge passed we use the equation:

I=\frac{q}{t}

where,

I = current passed = 1 A

q = total charge = ?

t = time required = 1 sec

Putting values in above equation, we get:

1A=\frac{q}{1s}\\\\q=1A\times 1s=1C

When 96500C of electricity is passed , the electrons passed = 6.022\times 10^{23}

1 C of electricity is passed , the electrons passed = \frac{6.022\times 10^{23}}{96500}\times 1C=6.241\times 10^{18}

Hence, 6.241\times 10^{18} electrons pass through a cross section of the wire each second.

You might be interested in
*NEED ANSWER QUICKLY!*
umka21 [38]
You would have to put the number together
8 0
3 years ago
The temperature at one of the Viking sites on Mars was found to vary daily from -90.OF to -5.0 C. Convert these temperatures to
kykrilka [37]

Answer:

I don't know about this, this is which class question

6 0
3 years ago
A prismatic bar AB of length L and solid circular cross section (diameter d) is loaded by a distributed torque of constant inten
Lyrx [107]

Answer:

a) the maximum shear stress τ_{max} the bar is 16T_{max} /πd³

b) the angle of twist between the ends of the bar is 16tL² / πGd⁴  

Explanation:

Given the data in the question, as illustrated in the image below;

d is the diameter of the prismatic bar of length AB

t is the intensity of distributed torque

(a) Determine the maximum shear stress tmax in the bar

Maximum Applied torque  T_max = tL

we know that;

shear stress τ = 16T/πd³

where d is the diameter

so

τ_{max} = 16T_{max} /πd³

Therefore, the maximum shear stress τ_{max} the bar is 16T_{max} /πd³

(b) Determine the angle of twist between the ends of the bar.

let theta (\theta) be the angle of twist

polar moment of inertia I_p} = πd⁴/32

now from the second image;

lets length dx which is at distance of "x" from "B"

Torque distance x

T(x) = tx

Elemental angle twist = d\theta = T(x)dx / GI_{p}

so

d\theta = tx.dx / G(πd⁴/32)

d\theta = 32tx.dx / πGd⁴

so total angle of twist \theta will be;

\theta =  \int\limits^L_0  \, d\theta

\theta =  \int\limits^L_0  \, 32tx.dx / πGd⁴

\theta = 32t / πGd⁴  \int\limits^L_0  \, xdx

\theta = 32t / πGd⁴ [ L²/2]

\theta = 16tL² / πGd⁴  

Therefore,  the angle of twist between the ends of the bar is 16tL² / πGd⁴  

7 0
3 years ago
The length of a certain wire is kept same while its radius is doubled. what is the new resistivity of this wire?
anastassius [24]
The text does not specify whether the resistance R of the wire must be kept the same or not: here I assume R must be kept the same.

The relationship between the resistance and the resistivity of a wire is
\rho =  \frac{AR}{L}
where
\rho is the resistivity
A is the cross-sectional area
R is the resistance
L is the wire length

the cross-sectional area is given by
A=\pi r^2
where r is the radius of the wire. Substituting in the previous equation ,we find
\rho =  \frac{\pi r^2 R}{L}

For the new wire, the length L is kept the same (L'=L) while the radius is doubled (r'=2r), so the new resistivity is
\rho' =  \frac{\pi r'^2 R}{L'}= \frac{\pi (2r)^2 R}{L}=4  \frac{\pi r^2 R}{L}   = 4 \rho
Therefore, the new resistivity must be 4 times the original one.
5 0
3 years ago
Read 2 more answers
During a particular thunderstorm, the electric potential difference between a cloud and the ground is vcloud - vground = 3.50 10
wariber [46]
<span>The change in the electron's potential energy is equal to the work done on the electron by the electric field. The electron's potential energy is the stored energy relative to the electron's position in the electric field. Vcloud - Vground represents the change in Voltage. This voltage quantity is given to be 3.50 x 10^8 V, with the electron at the lower potential. The formula for calculating the change in the electron's potential energy (EPE) is found by charge x (Vcloud - Vground) = (EPEcloud - EPE ground) where charge is constant = 1.6 x 10^-19. Filling in the known quantities results in the expression 1.6 x 10^-19 (3.50 x 10^8) = (EPEcloud - EPEground) = 5.6 x 10^-11. Therefore, the change in the electron's potential energy from cloud to ground is 5.6 x 10^-11 joules.</span>
8 0
3 years ago
Other questions:
  • A 2.50 gram rectangular object has measurements of 22.0 mm, 13.5 mm, and 12.5 mm. what is the object's density in units of g/ml?
    10·1 answer
  • If a bowling ball rolls 12 meters in 2.2 seconds what is the speed ?
    6·1 answer
  • For each situation, identify when sound would travel faster and why?
    11·1 answer
  • Why does increasing the heat under a pan of boiling water not decrease the cooking time for hard-boiled eggs? (Select all that a
    9·1 answer
  • I need help with this too. (im not good at science or math)
    8·2 answers
  • In what direction or orientation did the man throw the ball in image A? image B?​
    10·1 answer
  • What happens as lakes get older?
    12·1 answer
  • How are babies formed?​
    13·2 answers
  • When air gets hot, it rises and forms…
    14·1 answer
  • for hundreds of years scientists deny the existence of rogue waves until the presence of when was finally caught on record. when
    15·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!