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NARA [144]
3 years ago
8

A length of metal wire has a radius of 0.003 m and a resistance of 0.1 ω. when the potential difference across the wire is 16 v,

the electron drift speed is found to be 0.000282 m/s. based on these data, calculate the density of free electron in the wire. the fundamental charge is 1.6 × 10−19
c. answer in units of m−3 .
Physics
1 answer:
DENIUS [597]3 years ago
8 0
By definition;
Vd = I/nQA

In which,
Vd = drift speed = 0.000282 m/s
I = current through the wire = V/R = 16/0.1 = 160 Amps
n = Density of free electrons
Q = Fundamental charge = 1.6*10^-19 C
A = Cross-sectional area of the wire = πr^2 =π*0.003^2 = 2.827*10^-5 m^2

Rearranging the expression for Vd above and substituting;
n = I/VdQA = 160/(0.000282*1.6*10^-19*2.827*10^-5) = 1.254*10^29 free electrons/m^3
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7. How long does it take a ball rolling down a hill to change its speed from 3 m/sec to 34.5 m/sec
lys-0071 [83]

The time elapsed is 9 seconds

Explanation:

The motion of the ball is a uniformly accelerated motion (a motion with constant acceleration), so we can use the following suvat equation:

v=u+at

where :

v is the final velocity  of the ball

u is the initial velocity

a is the acceleration

t is the time  elapsed

For the ball in this problem, we have:

u = 3 m/s is the initial velocity

v = 34.5 m/s is the final velocity

a=3.5m/s^2 is the acceleration

Solving for t, we find the time taken for this change in velocity:

t=\frac{v-u}{a}=\frac{34.5-3}{3.5}=9 s

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8 0
3 years ago
Which area on the line has both some kinetic energy and some gravitational energy?
Yakvenalex [24]
I believe it is either A or B
4 0
2 years ago
Read 2 more answers
A heavy rope, 60 ft long, weighs 0.5 lb/ft and hangs over the edge of a building 130 ft high. (Let x be the distance in feet bel
Nataliya [291]

Answer:

Riemann sum

W = lim n→∞ Σ 0.5xᵢΔx (with the summation done from i = 1 to n)

Integral = W = ∫⁶⁰₀ 0.5x dx

Workdone in pulling the entire rope to the top of the building = 900 lb.ft

Riemann sum for pulling half the length of the rope to the top of the building

W = lim n→∞ Σ 0.5xᵢΔx (but the sum is from i = 1 to n/2)

Integral = W = ∫⁶⁰₃₀ 0.5x dx

Work done in pulling half the rope to the top of the building = 675 lb.ft

Step-by-step explanation:

Using Riemann sum which is an estimation of area under a curve

The portion of the rope below the top of the building from x to (x+Δx) ft is Δx.

The weight of rope in that part would be 0.5Δx.

Then workdone in lifting this portion through a length xᵢ ft would be 0.5xᵢΔx

So, the Riemann sum for this total work done would be

W = lim n→∞ Σ 0.5xᵢΔx (with the summation done from i = 1 to n)

The Riemann sum can easily be translated to integral form.

In integral form, with the rope being 60 ft long, we have

W = ∫⁶⁰₀ 0.5x dx

W = [0.25x²]⁶⁰₀ = 0.25 (60²) = 900 lb.ft

b) When half the rope is pulled to the top of the building, 60 ft is pulled up until the length remaining is 30 ft

Just like in (a)

But the Riemann sum will now be from the start of the curve, to it's middle

Still W = lim n→∞ Σ 0.5xᵢΔx (but the sum is from i = 1 to n/2)

W = ∫⁶⁰₃₀ 0.5x dx

W = [0.25x²]⁶⁰₃₀ = 0.25 (60² - 30²) = 675 lb.ft

Hope this Helps!!!

7 0
3 years ago
Just help me do this dawg pls
givi [52]

Answer:

D

Series means one after another like cars on a train

3 0
2 years ago
a point charge q1 is held stationary at the origin. a second charge q2 is placed at point a, and the electric potential energy o
Gnoma [55]

The electric potential energy of the pair of charges when the second charge is at point b is U at b = +7.3×10⁻⁸J.

Electric potential is the amount of work done in moving point charge from infinity to aspecific point.

Calculation of the electric potential energy:

Here the particle should be moved from the point at the time when the potential energy should be U a

The U b should be the change in the potential energy

W is work done.

(W at a )-(W at b)  = U a - U b

U b = U a - (W at a )+(W at b)

So,

= 5.4 × 10⁻⁸J - ( -1.9×10⁻⁸ J)

= +7.3×10⁻⁸J.

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