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dalvyx [7]
3 years ago
6

Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of -2.0 µC; sphere B carries a charge of -6.0 µC;

and sphere C carries a charge of +4.0 µC. Spheres A and B are touched together and then separated. SpheresB and C are then touched and separated. Does sphere C end up with an excess or a deficiency of electrons and how many electrons is it?Select one:A. deficiency, 6 × 10^13B. excess, 2 × 10^13C. There is no excess or deficiency of electrons.D. excess, 3 × 10^13E. deficiency, 3 × 10^12
Physics
1 answer:
Dmitry_Shevchenko [17]3 years ago
5 0

Answer:

None of the option is correct

A=-4µC, B=0 C, C=0 C

C will be +4.0 µC deficient after the contact

Explanation:

A and B are come in contact together, the charge will flow to establish  equilibrium, and hence becoming: A=-4µC, B=-4µC, C=+4.0 µC

Similarly when C and B touch, the positive and the negative will exact the same force due to equal charge magnitude and become electrically neutral : A=-4µC, B=0 C, C=0 C.

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kipiarov [429]
4-C. ... 5-D. ... 6-A. ... 7-D. ... 8-C. ... 9-B. ... 10-D. It's really a joy and a delight that you've learned so much by posting all of these questions.
5 0
4 years ago
A non uniform rod has mass
Doss [256]

Answer:

r_{cm} = L/3

Explanation:

Mass: M, Length: L.

\sigma (x) = b(L-x)

The formula that gives center of mass is

\vec{r}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + ...}{m_1 + m_2 + ...} = \frac{\Sigma m_i \vec{r}_i}{\Sigma m_i}

In the case of a non-uniform mass density, this formula converts to

\vec{r}_{cm} = \frac{\int\limits^L_0 {x\sigma(x)} \, dx }{\int\limits^L_0 {\sigma(x)} \, dx }

where the denominator is the total mass and the nominator is the mass times position of each point on the rod.

We have to integrate the mass density over the total rod in order to find the total mass. Likewise, we have to integrate the center of mass of each point (xσ(x)) over the total rod. And if we divide the integrated center of mass to the total mass, we find the center of mass of the rod:

\vec{r}_{cm} = \frac{\int\limits^L_0 {x\sigma(x)} \, dx }{\int\limits^L_0 {\sigma(x)} \, dx } = \frac{\int\limits^L_0 {xb(L-x)} \, dx }{\int\limits^L_0 {b(L-x)} \, dx } = \frac{b\int\limits^L_0{(xL - x^2)} \, dx }{b\int\limits^L_0 {(L-x)} \, dx } = \frac{\frac{x^2L}{2} - \frac{x^3}{3}}{Lx - \frac{x^2}{2}}\left \{ {{x=L} \atop {x=0}} \right.

Here x's are cancelled. Otherwise, the denominator would be zero.

r_{cm} = \frac{\frac{xL}{2}-\frac{x^2}{3}}{L-\frac{x}{2}}\left \{ {{x=L} \atop {x=0}} \right. = \frac{\frac{L^2}{2}-\frac{L^2}{3}}{L-\frac{L}{2}} = \frac{\frac{L^2}{6}}{\frac{L}{2}} = \frac{L}{3}

8 0
3 years ago
Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 90.0 kg, down a p slope at constant speed, as sho
kondor19780726 [428]

The work done by friction to move the sled is  - 1,323 J.

<h3>What is Coefficient of friction?</h3>
  • The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them.
  • Typically, it is represented by the Greek letter µ. In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.
  • The coefficient of friction has no dimensions because both F and N are measured in units of force (such as newtons or pounds). For both static and kinetic friction, the coefficient of friction has a range of values.
  • When an object experiences static friction, the frictional force resists any applied force, causing the object to stay at rest until the static frictional force is removed. The frictional force opposes an object's motion in kinetic friction.

Solution:

Given that

Coefficient of friction (µ) = 0.10

Mass (m) = 90kg

distance covered (d) = 30m

We use the formula:

friction work = -µmgdcos∅

friction work = -0.100 × 90 kg × 9.8 m/s² × 30 m × cos 60°

friction work = - 1,323 J

Know more about Coefficient of friction numerical brainly.com/question/19308401

#SPJ4

5 0
2 years ago
How many neutrons are in calcium
vladimir1956 [14]

Answer:

I believe there are 20 Neutrons in Calcium.

Explanation:

6 0
4 years ago
Answers to question 2,3,4
Sati [7]

Answer:

OMG

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Explanation:

5 0
3 years ago
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