A rough surface for infrared waves may be polished for

Physics

1 answer:

zloy xaker [14]2 years ago

7 0

Answer:

Radio Waves

Explanation:

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two boxes sit on a frictionless surface and are in contact with one another. the first box has a mass of 7 kg and the second box

egoroff_w [7]

The acceleration of the boxes depends on the mass and weight.

we have a mass of 7 and 8 kilograms

if it took 25 N force to move box A, then you would take 25 and multiply by 8 then divide by 2.

It will leave you with 100 N.

finally take the sq rt of 100 to get 10

7 0

3 years ago

Learning Goal:

enot [183]

Answer:

A. $U_0 = \dfrac{\epsilon_0 A V^2}{2d}$

B. $U_1 = \dfrac{\epsilon_0 A V^2}{6d}$

C. $U_2 = \dfrac{K\epsilon_0 A V^2}{2d}$

Explanation:

The capacitance of a capacitor is its ability to store charges. For parallel-plate capacitors, this ability depends the material between the plates, the common plate area and the plate separation. The relationship is

$C=\dfrac{\epsilon A}{d}$

$C$ is the capacitance, $A$ is the common plate area, $d$ is the plate separation and $\epsilon$ is the permittivity of the material between the plates.

For air or free space, $\epsilon$ is $\epsilon_0$ called the permittivity of free space. In general, $\epsilon=\epsilon_r \epsilon_0$ where $\epsilon_r$ is the relative permittivity or dielectric constant of the material between the plates. It is a factor that determines the strength of the material compared to air. In fact, for air or vacuum, $\epsilon_r=1$ .

The energy stored in a capacitor is the average of the product of its charge and voltage.

$U = \dfrac{QV}{2}$

Its charge, $Q$ , is related to its capacitance by $Q=CV$ (this is the electrical definition of capacitance, a ratio of the charge to its voltage; the previous formula is the geometric definition). Substituting this in the formula for $U$ ,