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12345 [234]
3 years ago
13

three concentric conducting spherical shells have radii a, b, c such that x< b (a) Find the electric potential of the three s

hells.
(b) If the inner and outer shells are now connected by a wire that is insulated as it passes through the middle shell, what is the electric potential of each of the three shells, and what is the final charge on each shell?
Physics
1 answer:
qwelly [4]3 years ago
4 0

Answer:

a) V_{A} =\frac{KQ(c-b)}{bc}

V_{B} =\frac{KQ(c-b)}{bc}

Vc = 0

b) qA = 0

qB = Q

qC = -Q

VA = VC = 0

V_{B} =\frac{KQ(c-b)}{bc}

Explanation:

a) The electric potential of the three shells are:

V_{A} =\frac{KQ}{b} +\frac{K(-Q)}{c} =KQ(\frac{1}{b} -\frac{1}{c} )=\frac{KQ(c-b)}{bc}

V_{B} =\frac{KQ}{b} +\frac{K(-Q)}{c} =KQ(\frac{1}{b} -\frac{1}{c} )=\frac{KQ(c-b)}{bc}

V_{c} =\frac{KQ}{c} +\frac{K(-Q)}{c} =0

b) When the inner shell is connected with another, both will have the same potential, then:

qA = 0

qB = Q

qC = -Q

V_{a} =V_{c}\frac{KQ}{c} +\frac{K(-Q)}{c} =0

V_{B} =\frac{KQ}{b} +\frac{K(-Q)}{c} =KQ(\frac{1}{b} -\frac{1}{c} )=\frac{KQ(c-b)}{bc}

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8. A rectangle is measured to be 6.4 +0.2 cm by 8.3 $0.2 cm.
mamaluj [8]

Answer:

a) The perimeter of the rectangle is 29.4 centimeters.

b) The uncertainty in its perimeter is 0.8 centimeters.

Explanation:

a) From Geometry we remember that the perimeter of the rectangle (p), measured in centimeters, is represented by the following formula:

p = 2\cdot (w+l) (1)

Where:

w - Width, measured in centimeters.

l - Length, measured in centimeters.

If we know that w = 6.4\,cm and l = 8.3\,cm, then the perimeter of the rectangle is:

p = 2\cdot (6.4\,cm+8.3\,cm)

p = 29.4\,cm

The perimeter of the rectangle is 29.4 centimeters.

b) The uncertainty of the perimeter (\Delta p), measured in centimeters, is estimated by differences. That is:

\Delta p = 2\cdot (\Delta w + \Delta l)  (2)

Where:

\Delta w - Uncertainty in width, measured in centimeters.

\Delta l - Uncertainty in length, measured in centimeters.

If we know that \Delta w = 0.2\,cm and \Delta l = 0.2\,cm, then the uncertainty in perimeter is:

\Delta p = 2\cdot (0.2\,cm+0.2\,cm)

\Delta p = 0.8\,cm

The uncertainty in its perimeter is 0.8 centimeters.

5 0
3 years ago
How do you find weight =mass×gravitational acceleration
Artist 52 [7]
That's a formula that shows the relationship between three quantities ...
weight, mass, and acceleration.  If you know any two of them, then you
can use this formula to find the one you don't know.

Examples:

==>  I have a rock with 2 kilograms of mass.
        The gravitational acceleration on Earth is 9.8 m/s² .
        How much does my rock weigh on Earth ?

         Weight = (mass) x (grav acceleration)
                      = (2 kg) x (9.8 m/s²)
                                                       =  19.6 newtons
                                                  (about 4.41 pounds)

==>  My brother weighs 770 newtons (about 173 pounds) on Earth.
         What is his mass ?

                               Weight = (mass) x (grav acceleration)

                               770 newtons = (mass) x (9.8 m/s²)
Divide each side
by 9.8 m/s²:           770 newtons / 9.8 m/s² = mass

                                        78.57 kilograms = mass

==> When I went to the Moon, I took along my 2-kilogram rock.
         I weighed my rock on the Moon. 
         It weighs 3.25 newtons  (about 0.73 pound) there.
         What is the gravitational acceleration on the Moon ?

                                   Weight = (mass) x (grav acceleration)

                                   3.25 newtons = (2 kg) x (acceleration)

Divide each side
by  2 kilograms:        (3.25 newtons)/(2 kg)  =  acceleration

                                    1.63 m/s² = grav acceleration on the Moon 

          


7 0
3 years ago
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3 years ago
A force of 6.0 N is applied horizontally to a 3.0 kg crate initially at rest on a horizontal frictionless surface. After the cra
Savatey [412]

Answer:

a. Yes, because the acceleration of the crate is 2.0 m/s².

Explanation:

Given

Force = 6N --- f

Mass = 3kg --- m

Time = 1.5s --- t

Velocity = 3.0m/s --- v

Required

Does the system support F=ma

Yes, it does and this is shown below

The crate is initially at rest; so:

u = 0

Using the first equation of motion

v = u + at

Substitute values for v, u and t

3 = 0 + a*1.5

3 = 1.5a

Make a the subject

a = 3/1.5

a = 2

Using F = ma

Substitute values for F and m

6 = 3 * a

Divide both sides by 3

6/3 = 3/3 * a

2 = a

a = 2

In both cases:

a = 2

<em>Hence, option (a) is correct.</em>

6 0
3 years ago
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