1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
GrogVix [38]
3 years ago
8

In the Faraday pail experiment a metal ball is lowered into a brass pail. In a variation of this experiment, suppose we charge t

he metal ball with Q and the brass pail with charge 2Q. The ball is slowly lowered into the pail. At no time does the sphere touch the pail. While the ball is inside, the charge on the outside of the pail is _____. The outer part of the pail is then touched/grounded. Then the ball is removed. After the ball is removed, the charge on the outer surface of the pail is ___. Briefly support your answers with reasoning from Gauss's Law.
Physics
1 answer:
sertanlavr [38]3 years ago
6 0

Answer:

a)

*The charges of which and the cube are of the same sign..

          Q_exterior = 3 Q

* the charge of the sphere has a different sign than the charge of the cube,

         Q_exterior =  Q

b)   Q = 0

Explanation:

To correctly describe this experiment, you must remember that in metals charges are mobile and that charges of the same sign repel and of different signs attract.

Let's analyze each situation

a) Suppose that the charges of which and the cube are of the same sign.

When the ball is introduced without touching the walls, its charge Q attracts a charge of equal magnitude and different sign to the internal wall. If we create a Gaussian surface around the inner wall of the sphere the net charge between the ball and the inner wall is zero. Consequently, a face Q should have been generated in the outer wall, therefore in this wall it has a total load of

               Q_exterior = 3 Q

   Now suppose that the charge of the sphere has a different sign than the charge of the cube, for simplicity let's say that the charge of the sphere is -Q and the cube + 2Q,

     Again we create a Gaussian surface outside the inner wall, now the charge on the ball attracts a charge of value + Q to neutralize the charge between the ball and the inner wall. Therefore a load remains on the outer wall

              Q_exterior = + Q

b) The cube is connected to earth and it is touched with the ball, in this case the charge of the two bodies is neutralized by the Earth, therefore the bodies have zero charge

             Q = 0

You might be interested in
The greatest speed with which an athlete can jump vertically is around 5 m/sec. Determine the speed at which Earth would move do
katrin2010 [14]

Answer:

Approximately 2.0 \times 10^{-23}\; \rm m \cdot s^{-1} if that athlete jumped up at 1.8\; \rm m \cdot s^{-1}. (Assuming that g = 9.81\; \rm m\cdot s^{-1}.)

Explanation:

The momentum p of an object is the product of its mass m and its velocity v. That is: p = m \cdot v.

Before the jump, the speed of the athlete and the earth would be zero (relative to each other.) That is: v(\text{athlete, before}) = 0 and v(\text{earth, before}) = 0. Therefore:

\begin{aligned}& p(\text{athlete, before}) = 0\end{aligned} and p(\text{earth, before}) = 0.

Assume that there is no force from outside of the earth (and the athlete) acting on the two. Momentum should be conserved at the instant that the athlete jumped up from the earth.

Before the jump, the sum of the momentum of the athlete and the earth was zero. Because momentum is conserved, the sum of the momentum of the two objects after the jump should also be zero. That is:

\begin{aligned}& p(\text{athlete, after}) + p(\text{earth, after}) \\ & =p(\text{athlete, before}) + p(\text{earth, before}) \\ &= 0\end{aligned}.

Therefore:

p(\text{athelete, after}) = - p(\text{earth, after}).

\begin{aligned}& m(\text{athlete}) \cdot v(\text{athelete, after}) \\ &= - m(\text{earth}) \cdot v(\text{earth, after})\end{aligned}.

Rewrite this equation to find an expression for v(\text{earth, after}), the speed of the earth after the jump:

\begin{aligned} &v(\text{earth, after}) \\ &= -\frac{m(\text{athlete}) \cdot v(\text{athlete, after})}{m(\text{earth})} \end{aligned}.

The mass of the athlete needs to be calculated from the weight of this athlete. Assume that the gravitational field strength is g = 9.81\; \rm N \cdot kg^{-1}.

\begin{aligned}& m(\text{athlete}) = \frac{664\; \rm N}{9.81\; \rm N \cdot kg^{-1}} \approx 67.686\; \rm N\end{aligned}.

Calculate v(\text{earth, after}) using m(\text{earth}) and v(\text{athlete, after}) values from the question:

\begin{aligned} &v(\text{earth, after}) \\ &= -\frac{m(\text{athlete}) \cdot v(\text{athlete, after})}{m(\text{earth})} \\ &\approx -2.0 \times 10^{-23}\; \rm m \cdot s^{-1}\end{aligned}.

The negative sign suggests that the earth would move downwards after the jump. The speed of the motion would be approximately 2.0 \times 10^{-23}\; \rm m \cdot s^{-1}.

3 0
3 years ago
PLEASE I NEED HELP I AM REALLY STUCK!!!
AlekseyPX

Answer: Force = Mass X Acceleration

F = 5 x 2

F = 10 N

8 0
3 years ago
A 1000-kg car moving at 10 m/s brakes to a stop in 5 s. the average braking force is
Marina86 [1]
To calcculate the braking force of the car moving, we use Newton's second law of motion which relates the acceleration and the force of an object moving. The force of an object moving is directly proportional to its acceleration and the proportionality constant is the mass of the object. It is expressed as:

Force = ma

Acceleration is the rate of change of the velocity of a moving object. We calculate acceleration from the velocity and the time given above.

a = (10 m/s) / 5 s = 2 m/s^2

So,
Force = ma
Force = 1000 kg ( 2 m/s^2 )
Force = 2000 kg m/s^2 or 2000 N
4 0
3 years ago
A 100 kg mass is pulled along a frictionless surface by a horizontal force F such that its acceleration is 8.0 m/s2. A 20 kg mas
nadya68 [22]

Answer:

a) 60 N

b) 860 N

Explanation:

Given that,

m_1 = 100 kg

m_2 = 20 kg

a_{1} = 8.0 \frac{m}{s^{2}}

a_{2} = 3.0 \frac{m}{s^{2}}

a) By Newton's Law,

∑F_{m_2,x} = f_k = m_{2} * a_{2}

∑F_{m_2,y} = F_{N} - m_{2} * g = 0

Hence,

f_k = m_2 * a_2 = 20 * 3 = 60 N

b) By Newton's Law

∑F_{m_1,x} = F = m_{1} * a_{1}

Hence,

F = m_{1} * a_{1} = 100 * 8 = 800 N

Net force acting on 100 kg mass,

F_{net} = F + f_k = 800 + 60 = 860 N

6 0
3 years ago
a car traveling north with a velocity of 29m/s changes its velocity to 36 in the span of 11s. What was the car's acceleration du
Nikolay [14]

Answer:

I’m so sorry I tried solving it but I don’t understand it can you explain the question a little bit more ty

Explanation:

8 0
4 years ago
Other questions:
  • A small rock is thrown vertically upward with a speed of 22.0 m/s from the edge of the roof of a 30.0-m-tall building. The rock
    6·1 answer
  • Which of the following does not exist in a command economy
    8·1 answer
  • Who discovered the strong nuclear force
    13·2 answers
  • What does the universal law of gravitation state
    14·2 answers
  • That which travels in waves that are not the same length is called
    15·1 answer
  • I need help with this work
    8·2 answers
  • How did puerto rico pose a problem for the united states after the spanish american wat
    9·1 answer
  • Can I have some help
    15·1 answer
  • Find the right answer please
    6·1 answer
  • Lightning flashes and you hear a thunderclap 4 seconds later. The velocity of sound is 340 m/s. How far away did the lightning s
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!