1. Ano-ano ang mga mabubuti at masasamang makataong kilos na nabanggit sa sitwasyon?

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

26. The speed of light in a certain medium is

horrorfan [7]

The number we need in order to answer the question belongs in the space between the words "is" and "of". You left that blank blank, so there really isn't any question here to answer.

HOWEVER ... the refractive index of a medium can never be less than 1.0 , so we know for sure that choice-a can't be the correct answer.

6 0

3 years ago

Gasoline burns in the cylinder of an automobile engine. During the combustion reaction, the production of gas forces the piston

serg [7]

Answer:

$\Delta U = 1640 J$

Explanation:

As we know by first law of thermodynamics that for ideal gas system we have

Heat given = change in internal energy + Work done

so here we will have

Heat given to the system = 2.2 kJ

Q = 2200 J

also we know that work done by the system is given as

$W = 560 J$

so we have

$\Delta U = Q - W$

$\Delta U = 2200 - 560$

$\Delta U = 1640 J$

6 0

3 years ago

The momentum of a falling rock is found to be 200 kg m/s. What is the mass of the rock if it falls with a velocity of 5.0 m/s

Snezhnost [94]

Answer:

$\boxed {\boxed {\sf 40 \ kilograms}}$

Explanation:

Momentum is the product of velocity and mass. The formula is:

$p=m*v$

We know the rock is falling. Its momentum is 200 kilograms meters per second and its velocity is 5 meters per second. Substitute the values into the formula.

$200 \ kg \ m/s = m * 5.0 \ m/s$

We are solving for m, the mass. We must isolate the variable. It is being multiplied by 5 meters per second. The inverse of multiplication is division, so we divided both sides by 5.0 m/s.

$\frac{200 \ kg \ m/s}{5.0 \ m/s}=\frac{ m* 5.0 \ m/s }{5.0 \ m/s}$