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Mandarinka [93]
4 years ago
10

The outer surface of the cardboard center of a paper towel roll: a. cannot be a Gaussian surface since it is an insulator. b. is

a possible Gaussian surface. c. cannot be a Gaussian surface because it is not a closed surface. d. none of the other responses. e. cannot be a Gaussian surface because it encloses no charge.
Physics
1 answer:
zvonat [6]4 years ago
5 0

Option C

The outer surface of the cardboard center of a paper towel roll: cannot be a Gaussian surface because it is not a closed surface.

<h3><u>Explanation:</u></h3>

A Gaussian surface is a bounded surface, ie. an impounded plain, bounded by space which can be practiced in obtaining out the mass of flux of electric, gravitational, or any additional field.

It should be closed surface so that a precise estimation can be performed within points that are inside the cover, on the surface, and outside the cover is the prime part of this. In this problem with considering that paper towel roll is not a closed surface, it is not a gaussian surface.

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3 years ago
A negative charge of 20 x 10-6C and another charge of 15 x 10-6C are separated by as distance of 0.7 m.
denpristay [2]

Answer:

Approximately 5.5\; \rm N, assuming that the volume of these two charged objects is negligible.

Explanation:

Assume that the dimensions of these two charged objects is much smaller than the distance between them. Hence, Coulomb's Law would give a good estimate of the electrostatic force between these two objects regardless of their exact shapes.

Let q_1 and q_2 denote the magnitude of two point charges (where the volume of both charged object is negligible.) In this question, q_1 = 20 \times 10^{-6}\; \rm C  and q_2 = 15 \times 10^{-6}\; \rm C.

Let r denote the distance between these two point charges. In this question, r = 0.7\; \rm m.

Let k denote the Coulomb constant. In standard units, k \approx 8.98755\times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2}.

By Coulomb's Law, the magnitude of electrostatic force (electric force) between these two point charges would be:

\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\end{aligned}.

Substitute in the values and evaluate:

\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\\ &\approx 8.98755 \times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2} \\ &\quad \times 20\times 10^{-6}\; \rm C\\ &\quad \times 15\times 10^{-6}\; \rm C \\ &\quad \times \frac{1}{{(0.7\; \rm m)}^{2}}\\ &\approx 5.5\; \rm N \end{aligned}.

8 0
3 years ago
Um objeto de 4cm de altura está a 30cm de um espelho côncavo, cujo raio de curvatura tem valor absoluto de 20cm.
Shkiper50 [21]

a) The distance of the image from the mirror is 15 cm

b) The size of the image is -2 cm (inverted)

Explanation:

a)

We can solve this first part of the problem by applying the mirror equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length

p is the distance of the object from the mirror

q is the distance of the image from the mirror

For a mirror, the focal length is half the radius of curvature, R:

f=\frac{R}{2}

For this mirror, R = 20 cm, so its focal length is

f=\frac{20}{2}=+10 cm (positive for a concave mirror)

Here we also know:

p = 30 cm is the distance of the object from the mirror

So, by applying the equation, we can find q:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{10}-\frac{1}{30}=\frac{1}{15} \rightarrow q = 15 cm

b)

We can solve this part by using the magnification equation:

M=-\frac{y'}{y}=\frac{q}{p}

where

y' is the size of the image

y is the size of the object

q is the distance of the image from the mirror

p is the distance of the object from the mirror

Here we have:

q = 15 cm

p = 30 cm

y = 4 cm

Solving for y', we find the size of the image:

y'=-y\frac{q}{p}=-(4)\frac{15}{30}=-2 cm

and the negative sign means that the image is inverted.

#LearnwithBrainly

6 0
3 years ago
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