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3 years ago

Question 10 of 15

Physics

2 answers:

fiasKO [112]3 years ago

8 0

Answer:

B. and D.

Explanation:

phases of matter are solid liquid and gas

Lady_Fox [76]3 years ago

8 0

Answer:

b and d

Explanation:

phases are solid liquid and gas

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Parallel Plates Consider a very large conducting plate at potential V0 suspended a distance d above a very large grounded plane.

velikii [3]

Answer:

$V = \frac{V_0x}{d}$

Explanation:

Since the field lines are parallel and the electric field is uniform between two parallel plates, a test charge would experience the same force of attraction or repulsion no matter where it is located in the field,

I attached an image that could help to understand the representation of the field. The formula used to calculate it is given by,

$E= -\frac{\Delta V}{x}$ (1)

If we want to consider the change in Voltage with respect to the position then it would be,

$E= -\frac{dV}{dx}$

According to the information provided, the potential is $V_0$ and there is a distance d, therefore

$E= -\frac{V_0}{d}$ (2)

Taking equation (1) we can clear V, to what we have,

$\frac{dV}{dx} = -E$

$dV = -Edx$

Integrating,

$V= - \int Edx$

Substituting (2)

$V = -\int \frac{V_0}{d} dx$

$V = \frac{V_0x}{d}$

Where x is the height from the grounded plate.

8 0

3 years ago

Atomic numbers for atoms _____ in increments of _____ as you move across the periodic table of elements

astra-53 [7]

Increase in increments of 1

8 0

3 years ago

What is the time constant of a series circuit where the capacitor is 0.330μF and the resistor is 10Ω ?

PtichkaEL [24]

Answer:

$\tau=3.3*10^{-6}s$

Explanation:

Take at look to the picture I attached you, using Kirchhoff's current law we get:

$C*\frac{dV}{dt}+\frac{V}{R}=0$

This is a separable first order differential equation, let's solve it step by step:

Express the equation this way:

$\frac{dV}{V}=-\frac{1}{RC}dt$

integrate both sides, the left side will be integrated from an initial voltage v to a final voltage V, and the right side from an initial time 0 to a final time t:

$\int\limits^V_v {\frac{dV}{V} } =-\int\limits^t_0 {\frac{1}{RC} } \, dt$