What is the magnitude and direction of the electric field atradiaConsider a coaxial conducting cable consisting of a conductingrod of radius R1 inside of a thin-walled conducting shell of radius 2(both are infinite length). Suppose the inner rod hasradiusR1= 1.3 mm and outer shell has radiusR2= 10R1Ifthe net charge density on the center rod isq1= 3.4×10−12C/mand the outer shell isq2=−2q1,a.)What is the magnitude and direction of the electric field atradial distancer= 5R1from the center rod
Answers
Answer:
E = 9.4 10⁶ N / C, The field goes from the inner cylinder to the outside
Explanation:
The best way to work this problem is with Gauss's law
Ф = E. dA = qint / ε₀
We must define a Gaussian surface, which takes advantage of the symmetry of the problem. We select a cylinder with the faces perpendicular to the coaxial.
The flow on the faces is zero, since the field goes in the radial direction of the cylinders.
The area of the cylinder is the length of the circle along the length of the cable
dA = 2π dr L
A = 2π r L
They indicate that the distance at which we must calculate the field is
r = 5 R₁
r = 5 1.3
r = 6.5 mm
The radius of the outer shell is
r₂ = 10 R₁
r₂ = 10 1.3
r₂ = 13 mm
r₂ > r
When comparing these two values we see that the field must be calculated between the two housings.
Gauss's law states that the charge is on the outside of the Gaussian surface does not contribute to the field, the charged on the inside of the surface is
λ = q / L
Qint = λ L
Let's replace
E 2π r L = λ L /ε₀
E = 1 / 2piε₀ λ / r
Let's calculate
E = 1 / 2pi 8.85 10⁻¹² 3.4 10-12 / 6.5 10-3
E = 9.4 10⁶ N / C
The field goes from the inner cylinder to the outside
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A fully loaded, slow-moving freight elevator has a cab with a total mass of 1400 kg, which is required to travel upward 37 m in 3.6 min, starting and ending at rest. The elevator's counterweight has a mass of only 930 kg, so the elevator motor must help pull the cab upward. What average power is required of the force the motor exerts on the cab via the cable?
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2N
250N
5000N
50000N
Answers
Answer:
50000N
Explanation:
Force = mass × acceleration
= 2500 × 20
= 50000N
Answers
Answer:
a).
b).
c). Δ
Explanation:
ΔE=kinetic energy
a).
b).
c).
net work= EkA+EkB
Answers
Given:-
Speed = 36 km/hr
converting speed into m/s
Speed = 36*5/18
Speed = 10 m/s
t = 10 sec
By using the Formula
Distance = Speed * time
D = 10*10
D = 100 m
Hope it helps....
Answers
The kinetic energy of an object is given by:
KE = 0.5mv²
KE = kinetic energy, m = mass, v = speed
Given values:
KE = 0.161J, v = 2.33m/s
Plug in and solve for m:
0.161 = 0.5m(2.33)²
m = 0.059kg
Answers
Answer:
0.158 A.
Explanation:
Mass of gold deposited = 3 x 10^-3 kg
= 3 g
Molar mass = 196 g/mol
Number of moles = 3/196
= 0.0153 mol.
Faraday's constant,
1 coloumb = 96500 C/mol
Quantity of charge, Q = 96500 * 0.0153
= 1477.04 C.
Remember,
Q = I * t
t = 2.59 hr
= 2.59 * 3600 s
= 9324 s
Current, I = 1477.04/9324
= 0.158 A.
Answer:
0.158A
Explanation:
Using Faraday's first law of electrolysis which states that the mass(m) of a substance deposited or liberated at any electrode is directly proportional to the quantity of charge or electricity (Q) passed. i.e
m ∝ Q
m = Z Q
Where;
Z is the proportionality constant called electrochemical equivalent.
Faraday also observed that when 1 Faraday of electricity is equivalent to 96500C of charge.
Also,
Quantity of charge (Q), which is the product of current (I) passing through and the time taken (t) for the electrolysis, is given by;
Q = I x t; ----------------------(i)
With all of these in place, now let's go answer the question.
Since the gold ions carry one elementary unit of positive charge, now let's write the cathode-half reaction for gold (Au) as follows;
Au⁺ + e⁻ = Au ---------------------(ii)
From equation (ii) it can be deduced that when;
1 Faraday (96500C) of electricity is passed, 1 mole of Au forms ( = 197 grams of Au) [molar mass of Au = 197g]
Then, 3.00 x 10⁻³ kg (= 3 g of Au) will be formed by 3g x 96500C / 197g = 1469.5C
Therefore, the quantity of charge (Q) deposited is 1469.5C
Substitute this value (Q = 1469.5C) and time t = 2.59h (= 2.59 x 3600 s) into equation (i);
Q = I x t
1469.5 = I x 2.59 x 3600
1469.5 = I x 9324
Solve for I;
I = 1469.5 / 9324
I = 0.158A
Therefore, the current in the cell during that period is 0.158A
Note:
1 mole of gold atoms = 176g
i.e the molar mass of gold (Au) is 176g
Answers
Answer:
Length
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Energy
Direct Current (DC)
Frequency
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Amount of substance
Luminous Intensity
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