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Vera_Pavlovna [14]
2 years ago
11

What is the relationship between resistance and current.

Physics
1 answer:
Tasya [4]2 years ago
7 0

Answer:

They are inversely proportional with each other. R ∝ \frac{1}{I}

Explanation:

By the equation V=IR, assume V is constant.

Since I and R are on the same side,

if I increase, R decreases, or the vice versa.

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Visible light
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Read 2 more answers
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,

U = \dfrac{CV^2}{2}

A. Substituting for C in U,

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

B. When the distance is 3d,

U_1 = \dfrac{\epsilon_0 A V^2}{2\times3d}

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

C. When the distance is restored but with a dielectric material of dielectric constant, K, inserted, we have

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

6 0
3 years ago
The flywheel of an engine has moment of inertia 2.50 kg m2 about its rotation axis. What constant torque is required to bring it
MrRissso [65]

Answer:

Explanation:

From the question we are told that

   The moment of inertia is  I = 2.50 \ kg \cdot m^2

    The final  angular speed is w_f =  400 rev/min  =  \frac{400 * 2\pi}{60}  = 41.89 \ rad/s

     The time taken is  t =  8.0 s

      The initial angular speed is  w_i  =  0\ rad/s

Generally the average angular acceleration is mathematically represented as

        \alpha  =  \frac{w_f - w_i }{t}

=>     \alpha  =  \frac{41.89}{8}

=>      \alpha  = 5.24 \ rad/s^2

Generally the torque is mathematically represented as

   \tau  =  I  *  \alpha

=>    \tau   =  5.24 *  2.50

=>     \tau   =  13.09 \  N \cdot m

5 0
3 years ago
A 1.5m long string weighs 0.0020 kg. It is tensioned to 100N. A disturbance travels along it with a wavelength of 1.5m, find:a)
Zigmanuir [339]

Answer:

the propagation velocity of the wave is 274.2 m/s

Explanation:

Given;

length of the string, L = 1.5 m

mass of the string, m = 0.002 kg

Tension of the string, T = 100 N

wavelength, λ = 1.5 m

The propagation velocity of the wave is calculated as;

v = \sqrt{\frac{T}{\mu} } \\\\\mu \ is \ mass \ per \ unit \ length \ of \ the \ string\\\\\mu = \frac{0.002 \ kg}{1.5 \ m} = 0.00133 \ kg/m\\\\v = \sqrt{\frac{100}{0.00133} } \\\\v = 274.2 \ m/s

Therefore, the propagation velocity of the wave is 274.2 m/s

7 0
2 years ago
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