CAN U HELP PLZZ? ;(

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

2 years ago

A jet airplane is in level flight. The mass of the airplane is m=9010kg. The airplane travels at a constant speed around a circu

Lyrx [107]

Answer:

The magnitude of the lift force L = 92.12 kN

The required angle is ≅ 16.35°

Explanation:

From the given information:

mass of the airplane = 9010 kg

radius of the airplane R = 9.77 mi

period T = 0.129 hours = (0.129 × 3600) secs

= 464.4 secs

The angular speed can be determined by using the expression:

ω = 2π / T

ω = 2 π/ 464.4

ω = 0.01353 rad/sec

The direction $\theta = tan^{-1} ( \dfrac{\omega ^2 R}{g})$

$\theta = tan^{-1} ( \dfrac{0.01353 ^2 \times (9.77\times 1609)}{9.81})$

θ = 16.35°

The magnitude of the lift force L = mg ÷ Cos(θ)

L = (9010 × 9.81) ÷ Cos(16.35)

L = 88388.1 ÷ 0.9596

L = 92109.32 N

L = 92.12 kN

3 0

2 years ago

An object takes 5 seconds to move 2 meters upward. How fast did it go?

jonny [76]

Answer:

2.5

Explanation:

5/2=2.5

8 0

2 years ago

Starting from rest, a basketball rolls from top of a hill to the bottom, reaching a translational speed of 6.8 m/s. Ignore frict

kkurt [141]

Answer:

Explanation:

for baseball

(a) Let the mass of the baseball is m.

radius of baseball is r.

Total kinetic energy of the baseball, T = rotational kinetic energy + translational kinetic energy

T = 0.5 Iω² + 0.5 mv²

Where, I be the moment of inertia and ω be the angular speed.

ω = v/r

T = 0.5 x 2/3 mr² x v²/r² + 0.5 mv²

T = 0.83 mv²

According to the conservation of energy, the total kinetic energy at the bottom is equal to the total potential energy at the top.

m g h = 0.83 mv²

where, h be the height of the top of the hill.

9.8 x h = 0.83 x 6.8 x 6.8

h = 3.93 m

(b) Let the velocity of juice can is v'.

moment of inertia of the juice can = 1/2mr²

So, total kinetic energy

T = 0.5 x I x ω² + 0.5 mv²

T = 0.5 x 0.5 x m x r² x v²/r² + 0.5 mv²

m g h = 0.75 mv²

9.8 x 3.93 = 0.75 v²

v = 7.2 m/s

7 0

3 years ago

A horizontal 2.00\ m2.00 m long, 5.00\ kg5.00 kg uniform beam that lies along the east-west direction is acted on by two forces.

Sunny_sXe [5.5K]

Answer: $240\ rad/s^2$

Explanation:

Given

Length of beam $l=2\ m$

mass of beam $m=5\ kg$

Two forces of equal intensity acted in the opposite direction, therefore, they create a torque of magnitude

$\tau =F\times l=200\times 2=400\ N.m$

Also, the beam starts rotating about its center

So, the moment of inertia of the beam is

$I=\dfrac{ml^2}{12}=\dfrac{5\times 2^2}{12}\\\\I=\dfrac{5}{3}\ kg.m^2$

Torque is the product of moment of inertia and angular acceleration

$\Rightarrow \tau=I\alpha\\\\\Rightarrow 400=\dfrac{5}{3}\times \alpha\\\\\Rightarrow \alpha =240\ rad/s^2$

7 0

2 years ago