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3 years ago

A negative test charge experiences a force to the right as a result of an electric field. Which is the best conclusion to draw

Physics

1 answer:

zaharov [31]3 years ago

7 0

Answer:

The electric field points to the left because the force on a negative charge is opposite to the direction of the field.

Explanation:

In an electric field, a positive charge has tendency to move from high to low potential and hence experience the electric force in the direction of electric field since electric field lines are directed from high to low potential.

In an electric field, a negative charge has tendency to move from low to high potential and hence experience the electric force in the direction opposite to electric field since electric field lines are directed from high to low potential.This phenomenon happened because The electric field from a positive charge will points away from the charge while the electric field from a negative charge will points toward the charge.

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How an elastic material can no longer be elastic but plastic instead?

Dvinal [7]

deformação elástica – é aquela em que removidos os esforços atuando sobre o corpo, ele volta a sua forma originaldeformação plástica – é aquela em que removidos os esforços, não há recuperação da forma original.

3 0

4 years ago

Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J =

Sloan [31]

Answer:

The current is $I = 8.9 *10^{-5} \ A$

Explanation:

From the question we are told that

The radius is $r = 3.17 \ mm = 3.17 *10^{-3} \ m$

The current density is $J = c\cdot r^2 = 9.00*10^{6} \ A/m^4 \cdot r^2$

The distance we are considering is $r = 0.5 R = 0.001585$

Generally current density is mathematically represented as

$J = \frac{I}{A }$

Where A is the cross-sectional area represented as

$A = \pi r^2$

=> $J = \frac{I}{\pi r^2 }$

=> $I = J * (\pi r^2 )$

Now the change in current per unit length is mathematically evaluated as

$dI = 2 J * \pi r dr$

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows

$I = 2\pi \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr$

$I = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr$

$I = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ] | \left 0.001585} \atop 0}} \right.$

$I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ]$

substituting values

$I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ]$

$I = 8.9 *10^{-5} \ A$

5 0

4 years ago

Explain how this increases their efficiency. Explain how this increases their efficiency. Antireflective coating causes the phas

sweet [91]

Answer:

c) True. If the coating cancels the light requested by the reflection, so there is more energy to enter the cell and therefore its efficiency increases

Explanation:

This exercise asks to analyze the effect of the antireflective coating on the efficiency of solar cells.

Let's start by expressing the expression for the interference of two light beams taken at when

* the phase change introduced when passing from air to 180º film

* the wavelength change by the refractive index of the film ln = lo / n

therefore the expression for destructive interference is

2 n t = m λ

where m is an integer

with these concepts we can analyze the different statements

a) False. Phase shift does not change the wavelength of light

b) False. The refractive index of the solar cell is not affected by the refractive index of the film since the two materials do not mix.

c) True. If the coating cancels the light requested by the reflection, so there is more energy to enter the cell and therefore its efficiency increases

d) false. In solar cells the incidence is almost normal, therefore the effect of refraction (separation of colors for different angles) is very small

7 0

3 years ago

Henry mixed salt and water together in a cup until he observed a clear solution. He measured the mass of the solution. Then he p

Tomtit [17]

Answer:

B- He is correct. Evaporation is a physical change, but dissolving salt in water is a chemical change. The change in mass is evidence that a chemical change occurred.

Explanation: Dissolving Salt in water is a Chemical Change, Because the Salt arrangement is different in solid state than dissolved in water. As we can see in the image below, once the Salt is dissolved, it is separated into its ions, Na+ and Cl- Now, The evaporation process is a physical change, because the water doesn´t change its configuration H20 and it only changes its form, as we can see in the image below.

6 0

3 years ago

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What distinguishes a nebula and a star?

Kryger [21]

What distinguishes a nebula and a star is that a nebula is a cloud of gas and dust in outer space, visible in the night sky either as an indistinct bright patch or as a dark silhouette against other luminous matter. And a star is a type of astronomical object consisting of a luminous spheroid of plasma held together by its own gravity.

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4 years ago