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podryga [215]
3 years ago
13

Complete the following statement: When a glass rod is rubbed with silk cloth, the rod becomes positively charged asa) negative c

harges are transferred from the rod to the silk.b) negative charges are transferred from the silk to the rod.c) positive charges are created on the surface of the rod.d) positive charges are transferred from the silk to the rod.e) positive charges are transferred from the rod to the silk.
Physics
1 answer:
Ronch [10]3 years ago
4 0

Answer:

a) negative charges are transferred from the rod to the silk.

Explanation:

Only negative charge are transfered by electrons.

Rubbing the glass rod with the silk cause electrons to flow from the glass rod to the silk. This leaves the glass rod with a positive charge.

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3 years ago
If a participant were holding two different weights in their hands and the jnd for a 10-gram weight was 1 gram, what should the
Nataliya [291]

The jnd for a 100-gram weight, according to Weber's law will be 10 gram.

<h3>What is Weber's law?</h3>

It should be noted that Weber's law asserts that the nature of any given stimulus will always affect how change is perceived. In other words, the size, weight, importance, etc. of the prior situation and the significance of the change both influence whether a change will be observed.

In this case, it was given that the jnd for a 10-gram weight was 1 gram, therefore, the jnd for 100 gram will be;

= 100 / 10

= 10 gram

Therefore, jnd for a 100-gram weight, according to Weber's law will be 10 grams.

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7 0
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What is meant by rusting​
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Charge g is distributed in a spherically symmetric ball of radius a. (a) Evaluate the average volume charge density p. (b) Now a
nasty-shy [4]

Answer:

Explanation:

The volume of a sphere is:

V = 4/3 * π * a^3

The volume charge density would then be:

p = Q/V

p = 3*Q/(4 * π * a^3)

If the charge density depends on the radius:

p = f(r) = k * r

I integrate the charge density in spherical coordinates. The charge density integrated in the whole volume is equal to total charge.

Q = \int\limits^{2*\pi}_0\int\limits^\pi_0  \int\limits^r_0 {k * r} \, dr * r*d\theta* r*d\phi

Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0  \int\limits^r_0 {r^3} \, dr * d\theta* d\phi

Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0 {\frac{r^4}{4}} \, d\theta* d\phi

Q = k *\int\limits^{2*\pi}_0 {\frac{\pi r^4}{4}} \,  d\phi

Q = \frac{\pi^2 r^4}{2}}

Since p = k*r

Q = p*π^2*r^3 / 2

Then:

p(r) = 2*Q / (π^2*r^3)

3 0
3 years ago
A metal rod of length (L) moves with velocity (v), perpendicular to its length, in a magnetic field B, which is perpendicular to
Alborosie

Explanation:

If a metal rod of length L moves with velocity v is moving perpendicular to its length, in a magnetic field B, the induced emf is given by :

\epsilon=Blv

The electric field in the conductor is given by :

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It is clear that the electric field is independent of the length of the rod. If the length of the rod is doubled, the electric field in the rod remains the same.

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