Earth's atmosphere traps energy from the sun which is a direct result of the trapping of energy by Earth's atmosphere?

A coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged con

Tcecarenko [31]

Complete question:

A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).

Required:

What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.

Answer:

The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Explanation:

Given;

charge of the coaxial capable, Q = 8.5 µC = 8.5 x 10⁻⁶ C

length of the conductor, L = 50 m

inner radius, r₁ = 1.304 mm

outer radius, r₂ = 9.249 mm

The magnitude of the electric field halfway between the two cylindrical conductors is given by;

$E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}$

Where;

λ is linear charge density or charge per unit length

r is the distance halfway between the two cylindrical conductors

$r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \ \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm$

The magnitude of the electric field is now given as;

$E = \frac{8.5*10^{-6}}{2\pi(8.85*10^{-12})(5.2765*10^{-3})(50)} \\\\E = 5.793*10^5 \ V/m$

Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

5 0

3 years ago

In 1995 a research group led by Eric Cornell and Carl Wiemann at the University of Colorado successfully cooled Rubidium atoms t

saveliy_v [14]

Answer:

0.00493 m/s

Explanation:

T = Temperature of the isotope = 85 nK

R = Gas constant = 8.341 J/mol K

M = Molar mass of isotope = 86.91 g/mol

Root Mean Square speed is given by

$v_r=\sqrt{\dfrac{3RT}{M}}\\\Rightarrow v_r=\sqrt{\dfrac{3\times 8.314\times 85\times 10^{-9}}{86.91\times 10^{-3}}}\\\Rightarrow v_r=0.00493\ m/s$

The Root Mean Square speed is 0.00493 m/s

6 0

4 years ago

When you are ice skating, to get started, you push your skate backwards on the ice and, as a result, begin to move forward. Whic

Setler79 [48]

I would say the third... force.

3 0

3 years ago

Read 2 more answers

PLEASE HELP ME WITH THIS PROBLEM

valentinak56 [21]

1) The mass of the continent is $2.13\cdot 10^{21} kg$

2) The kinetic energy of the continent is 274.8 J

3) The speed of the jogger must be 2.76 m/s

Explanation:

1)

The continent is a slab of side 5900 km (so the surface is 5900 x 5900, assuming it is a square) and depth 26 km, therefore its volume is:

$V=(36)(4600)^2=7.62\cdot 10^8 km^3 = 7.62\cdot 10^{17} m^3$

The mass of the continent is given by

$m=\rho V$

where:

$\rho = 2790 kg/m^3$ is its density

$V=7.62\cdot 10^{17} m^3$ is its volume

Substituting, we find the mass:

$m=(2790)(7.62\cdot 10^{17})=2.13\cdot 10^{21} kg$

2)

To find the kinetic energy, we need to convert the speed of the continent into m/s first.

The speed is

v = 1.6 cm/year

And we have:

1.6 cm = 0.016 m

$1 year = (365)(24)(60)(60)=3.15\cdot 10^7 s$

So, the speed is

$v=\frac{0.016 m}{3.15 \cdot 10^7 s}=5.08\cdot 10^{-10}m/s$

Now we can find the kinetic energy of the continent, which is given by

$K=\frac{1}{2}mv^2$

where

$m=2.13\cdot 10^{21} kg$ is the mass

$v=5.08\cdot 10^{-10}m/s$ is the speed

Substituting,

$K=\frac{1}{2}(2.13\cdot 10^{21})(5.08\cdot 10^{-10})^2=274.8 J$

3)

The jogger in this part has the same kinetic energy of the continent, so

K = 274.8 J

And its mass is

m = 72 kg

We can write his kinetic energy as

$K=\frac{1}{2}mv^2$

where

v is the speed of the man

And solving the equation for v, we find his speed:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(274.8)}{72}}=2.76 m/s$

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

3 0

3 years ago

****PLEASE HELP**** WILL MARK BRAINLIEST

allochka39001 [22]

Answer:

The Current decreases

Explanation:

HOPE THIS HELPS!

4 0

3 years ago