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KATRIN_1 [288]
3 years ago
14

The diameter of a 12-gauge copper wire is 0.081 in. The maximum safe current it can 17) carry (in order to prevent fire danger i

n building construction) is 20 A. At this current, what is the drift velocity of the electrons?
Physics
1 answer:
Soloha48 [4]3 years ago
6 0

Answer:

0.44m/s

Explanation:

drift velocity=I/nAq

diameter 12 gauge

wire=0.081inches=0.081*2.5=0.2025cm radius=0.10125cm area=pi*R^2 =20/8.5*10^22*3.14*0.10125^2*10^-4*1.6*10^-19*

V = 0.44m/s

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Enter the expression
vlabodo [156]
I see the word "when..." kind of fading out at the end of the first line.
Whatever comes after it may be important.

If you're just supposed to copy the expression into the box,
then the problem is that you left the 'e' out of it.

I'm guessing that you're supposed to enter whatever the expression becomes
when either  N₀ or  ' t ' has some special value that's in the first line.

Just taking a wild guess here . . . . .

If it's  "Enter the expression ..... , when t=0 ." ,
then the correct answer in the box is     N₀  .

But that's just a wild guess.  As I pointed out, you cut off
the picture in the middle of the word 'when', and I've got
a hunch that there's something important after it.
6 0
3 years ago
A charged wire of negligible thickness has length 2L units and has a linear charge density λ. Consider the electric field E-vect
Stels [109]

Answer:

E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}

Explanation:

Given that

Length= 2L

Linear charge density=λ

Distance= d

K=1/(4πε)

The electric field at point P

E=2K\int_{0}^{L}\dfrac{\lambda }{r^2}dx\ sin\theta

sin\theta =\dfrac{d}{\sqrt{d^2+x^2}}

r^2=d^2+x^2

So

E=2K\lambda d\int_{0}^{L}\dfrac{dx }{(x^2+d^2)^{\frac{3}{2}}}

Now by integrating above equation

E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}

4 0
3 years ago
In levelling, the following staff readings were observed involving an inverted staff, A = 2.915 and B = -2.028. What is the rise
salantis [7]

Answer:

The rise from A to B is 0.887

Solution:

As per the question:

The following reading of an inverted staff is given as:

A = 2.915

B = -2.028

Here, for inverted staff, the greater reading shows greater elevation and lesser reading shows lower elevation.

Thus

The rise from A to B is given as:

A - B = 2.915 - 2.028 = 0.887

8 0
3 years ago
WILL UPVOTE!!!Physics help please!!
liubo4ka [24]
Speed v = initial speed u + acceleration a x time t 
v=u+at = 2 + 4*3 = 14 m/s

8 0
3 years ago
A spring has a spring constant of 90N/m.How much potential energy does it store when stretched by 2 cm?
il63 [147K]

Answer:

The potential energy stored in the spring is 0.018 J.

Explanation:

Given;

spring constant, k = 90 N/m

extension of the spring, x = 2 cm = 0.02 m

The potential energy stored in the spring is calculated as;

U = ¹/₂kx²

where;

U is the potential energy stored in the spring

Substitute the given values in the equation above;

U = ¹/₂  x  90 N/m  x  (0.02 m)²

U = 0.018 J

Therefore, the  potential energy stored in the spring is 0.018 J.

5 0
3 years ago
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