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Soloha48 [4]
4 years ago
11

An infinite line of charge with linear density λ1 = 6 μC/m is positioned along the axis of a thick insulating shell of inner rad

ius a = 2.6 cm and outer radius b = 4.2 cm. The insulating shell is uniformly charged with a volume density of rho = -634 μC/m3. What is λ2, the linear charge density of the insulating shell?
Physics
1 answer:
Anna11 [10]4 years ago
5 0

Answer: λ2= 2.34 * 10^-6 C/m

Explanation: In order to calculate the value of the  linear charge density of the insulating shell we have to multiply ρ* Volume of the hollow cylinder, so

Volume of cylinder:2*π*b*L *(b-a)  where (b-a) is the thickness, then

λ2=Q/L = 634 *10^-6 C/m^3* 2*π*0.042 m*(0.042-0.26)== 2.34 μ C/m

You might be interested in
A negative charge, q1, of 6 µC is 0. 002 m north of a positive charge, q2, of 3 µC. What is the magnitude and direction of the e
prohojiy [21]

Force on the particle is defined as the application of the force field of one particle on another particle. The magnitude and direction of the electrical force will be 4.05×10⁴N towards the north.

<h3>What is electrical force?</h3>

Force on the particle is defined as the application of the force field of one particle on another particle. It is a type of virtual force.

The given data in the problem is

q₁ is the negative charge = 6 µC=6×10⁻⁶ C

q₂ is the positive charge = 3 µC=3×10⁻⁶ C

r is the distance between the charges=0.002 m

F_E is the electric force =?

The value of electric force will be;

\rm F_E= \frac{Kq_1q_2}{r^2} \\\\ F_E= \frac{9\times 10^9\times 6\times 10^{-6}\times3\times10^{-6}}{(0.002)^2}\\\\ \rm F_E=4.05\times10^4\;N

Hence the magnitude and direction of the electrical force will be 4.05×10⁴N towards the north.

To learn more about the electrical force refer to the link;

brainly.com/question/1076352

7 0
2 years ago
A water tower is a familiar sight in many towns. The purpose of such a tower is to provide storage capacity and to provide suffi
Mademuasel [1]

The image of the water tower and the houses is in the attachment.

Answer: (a) P = 245kPa;

(b) P = 173.5 kPa

Explanation: <u>Gauge</u> <u>pressure</u> is the pressure relative to the atmospheric pressure and it is only dependent of the height of the liquid in the container.

The pressure is calculated as: P = hρg

where

ρ is the density of the liquid, in this case, water, which is ρ = 1000kg/m³;

When it is full the reservoir contains 5.25×10⁵ kg. So, knowing the density, you know the volume:

ρ = \frac{m}{V}

V = ρ/m

V = \frac{5.25.10^{5}}{10^{3}}

V = 525 m³

To know the height of the spherical reservoir, its diameter is needed and to determine it, find the radius:

V = \frac{4}{3}.\pi.r^{3}

r = \sqrt[3]{ \frac{3}{4\pi } .V}

r = \sqrt[3]{\frac{525.3}{4\pi } }

r = 5.005 m

diameter = 2*r = 10.01m

(a) Height for House A:

h = 15 + 10.01

h = 25.01

P = hρg

P = 25.01.10³.9.8

P = 245.10³ Pa or 245kPa

(b) h = 25 - 7.3

h = 17.71

P = hρg

P = 17.71.1000.9.8

P = 173.5.10³ Pa or 173.5 kPa

4 0
3 years ago
The analysts at Techno InfoSystems are considering the four-model approach to system development for a new client.
garri49 [273]

​It will provide a clear picture of current system functions before any modifications or improvements are made is a benefit if they use the four-model approach.

<u>Option: D</u>

<u>Explanation:</u>

The four model approach is followed by number of analyst which showcase that they construct the physical and logical model of both current and new system. The most important advantage of such approach is it portrays transparent image of ongoing system, before one apply any modification or variation. This is necessary because the flaws which generated earlier in system may affect the SDLC phases and outcome of such process may result into unsatisfied user by paying additional cost. This can be avoided by taking additional steps which make it worth it. The major disadvantage of such approach is the added time and cost of constructing model in both current system.

4 0
3 years ago
Planet-X has a mass of 3.42×1024 kg and a radius of 8450 km. What is the First Cosmic Speed i.e. the speed of a satellite on a l
tino4ka555 [31]

Answer:

First cosmic speed = 5195.74m/s

Second cosmic speed = 7346.05m/s

The raduis of the synchronous 0rbit of satellite is 2.80×10^7m

Explanation:

The first cosmic speed Is determined using the Orbital speed equation which is given by:

V = Sqrt(GM/r)

Where G = gravitational constant = 6.67 ×10^-11

M = Mass of planet

r = radius of the planet

V = Sqrt (6.67×10^-11)(3.42×10^24)/(8450×10^3)

V = Sqrt (2.28×10^14)/(8450×10^3)

V = Sqrt ( 26995739.64)

V = 5195.75m/s

The second cosmic speed is given by :

V = Sqrt(2 × GM)/r

V = Sqrt (2 × (6.67×10^-11)(3.42×10^24)/(8450×10^3)

V = Sqrt( 4.5×10^14)/ (8450×10^3)

V = Sqrt(53964497.04)

V = 7346.05m/s

The raduis of the synchronous orbit if the satellite around the planet is given by:

r = Cuberoot(T^2GM/4 pi r^2where T is the period of rotation of the planet in second

Given :

T = 17.1 hours converting to seconds

T = 17.1 ×60 ×60 = 61560 seconds

Substituting into the equation

r = Cuberoot ([(61560)^2×(6.67×10^-11)(3.42×10^24)/ (4 ×3.142×r^2)]

r = 2.80×10^7m

5 0
3 years ago
Read 2 more answers
A rectangular painting measures 1.0 m tall along the y' axis and 3.0 m wide along the
Gennadij [26K]

Answer:

a) 1 m tall, 3 m wide

b) 1 m tall, 1.31 m wide

Explanation:

According to the captain of the spaceship, the dimensions of the picture is the same i.e 1.0 m tall along the y' axis and 3.0 m wide along the x' axis.

b) The dimensions of the picture as seen by an observer on the Earth along the y axis will remain the same, 1.0 m tall, for the direction of the y axis is perpendicular to the spaceship movement.

The dimensions of the picture as seen by an observer on the Earth along the x axis will reduce if we are to go by the Lorentz contraction:

L(x) = L(x)' * √[1 - (v²/c²)]

where

L(x)' = the dimensions of the picture along the x axis on the spaceship,

v² = the speed of the spaceship and c² = the speed of light in the vacuum.

On substituting, we have

L(x) = 3 * √[1 - (0.81c²/c²)]

L(x) = 1.31 m

8 0
3 years ago
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