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natali 33 [55]
3 years ago
12

A 1.50 m cylinder of radius 1.10 cm is made of a complicated mixture of materials. Its resistivity depends on the distance x fro

m the left end and obeys the formula rho( x ) = a + bx^2, where a and b are constants. At the left end, the resistivity is 2.25 10^-8 Omega*m, while at the right end it is 8.50 x 10^-8 Omega*m.
What is the resistance of this rod?

What is the electric field at its midpoint if it carries a 1.75 A current?

If we cut the rod into two 75.0 cm halves, what is the resistance of each half?
Physics
1 answer:
Semenov [28]3 years ago
8 0
I have solved parts A and B by doing the following:
A) R = ρL/A First I found the constants a and b by solving for them, which I found to be: a = 2.25x10^-8 Ωm b = 2.78x10^-8 Ωm A = πr^2 A = 3.8013x10-4 m2 Taking the integral of the Resistance equation gave me: 1/A*(ax + bx3/3) I then took the integral from 0 to 1.50m, thereby giving me: (1/3.801299x10-4)*3.38x10-8*1.50 + 2.78x10-8*1.53/3 = 1.71x10-4 This I know to be correct.
B) ∫E.dl = V dV/dx = d/dx(IR) 1.75/A*(a + bx2) Plugging in 0.75 for x gave me E = 1.76x10^-4 V/m This is also correct
C) This is where I'm stuck. I had thought that if I used the integral in part A to find the integral from 0 to 0.75 then that would be my resistance in the left hand side. Then I thought that I could subtract the left hand value from the value I found in part A to give me the right hand value, but this hasn't worked. My workings are as follow: R = ρL/A A = 3.8x10^-4 m2 Rleft hand1/A*(2.25x10^-8)(0.75) + (2.78x10^-8)(0.75^3/3) = 4.43966x10^-5 Then Rright hand = 1.71x10^-4 - 4.43966x10^-5 = 1.266x10^-4 
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A commuter backs her car out of her garage with an acceleration of 1.40 m/s2.A) How long does it take her to reach a speed of 2.
AURORKA [14]

Answer:

(A) 1.43secs

(B) -2.50m/s^2

Explanation:

A commuter backs her car out of her garage with an acceleration of 1.40m/s^2

(A) When the speed is 2.00m/s then, the time can be calculated as follows

t= Vf-Vo/a

The values given are a= 1.40m/s^2 , Vf= 2.00m/s, Vo= 0

= 2.00-0/1.40

= 2.00/1.40

= 1.43secs

(B) The deceleration when the time is 0.800secs can be calculated as follows

a= Vf-Vo/t

= 0-2.00/0.800

= -2.00/0.800

= -2.50m/s^2

4 0
2 years ago
Which of the following best describes the human population from early times to the present?
Makovka662 [10]

Answer:

We have a not significant increase of the population until 1700s or 1800s and then a significant increase growth from these years to the present.

Explanation:

From the figure attached we see the evolution of the human population since early times (1050).

We see that from 1050 until 1750-1850 we have an increase slowly with a low value for the increase per year.

But after these years (1750-1850) we see a considerable increase of the population, like an exponential model.

So then we can conclude in general terms this:

We have a not significant increase of the population until 1700s or 1800s and then a significant increase growth from these years to the present.

7 0
3 years ago
A block is pulled along a horizontal surface at a constant speed by a force (14.1i 0 j 5.1k). The direction k is perpendicular t
Pepsi [2]

Answer:

W =84.6\ Nm

Explanation:

given,

F = 14.1 i + 0 j + 5.1 k

displacement = 6 m

Assuming block is moving in x- direction

we know,

 dW = F dx

\int dW = F\int dx

W = F\int_0^6 dx

W = F[x]_0^6

W = 14.1 \times 6

W =84.6\ Nm

hence, work done by the force is equal to W =84.6\ Nm

7 0
3 years ago
A flat sheet is in the shape of a rectangle with sides of lengths 0.400 m and 0.600 m. The sheet is immersed in a uniform electr
Katen [24]

Answer:

ФE = 9.403W

Explanation:

In order to calculate the magnitude of the electric flux trough the sheet, you use the following formula:

\Phi_E=\vec{A}\cdot \vec{E}=AEcos\theta       (1)

A: area of the rectangular sheet = (0.400m)(0.600m) = 0.24m^2

E: magnitude of the electric field = 95.0N/C

θ: angle between the direction of the electric field and the normal to the surface of the sheet

You replace the values of the parameters in the equation (1):

\Phi_E=(0.24m^2)(95.0N/m)cos(20\°)=9.304W

The magnitude of the electric flux is trough the sheet is 9.403W

5 0
2 years ago
A charge of +3.0 mC is distributed uniformly along the circumference of a circle with a radius of 20 cm. How much external energ
Marat540 [252]

Answer:

Work done = 4584.9 J

Explanation:

given: q1=3.0 mC = 3.0 × 10⁻³ C, r = 20 cm = 0.20 m, q1 = 34μC = 34 × 10⁻⁶ C

Solution:

Formula for the potential difference at the center of the circle

P.E = K × q1 q2 /r   (Coulomb's constant k= 8.99 × 10⁹ N·m² / C²)

P.E = 8.99 × 10⁹ N·m² / C² × 3.0 × 10⁻³ C × 34 × 10⁻⁶ C /  0.20 m

P.E =  4584.9 J = Work done

3 0
3 years ago
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