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Kobotan [32]
2 years ago
7

An uncharged spherical conducting shell surrounds a charge –q at the center of the shell. Then charge +3q is placed on the outsi

de of the shell.
When static equilibrium is reached, the charges on the inner and outer surfaces of the shell are respectively:

a) +q, -q
b) -q, +q
c) +q, +2q
d) +2q, +q
Physics
1 answer:
Veronika [31]2 years ago
8 0

Answer:

a) The the charges on the inner and outer surfaces of the shell are respectively +q, -q

Explanation:

Under static equilibrium inside a conductor, the total electric field, E = 0

This must be zero so that no charge will be moving since the conductor is in static equilibrium.

Also, since Electric field, E is zero, then flux through the surface will zero.

From Gauss' law, the total charge enclosed is zero.

Given –q as the  charge at the center of the shell, then the opposite charge on inner surfaces  will be +q, so that the total charge enclosed will be zero.

Since the charge is in static equilibrium, then opposite charge will be on the surface, that is –q.

Therefore, the the charges on the inner and outer surfaces of the shell are respectively +q, -q

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13. An aircraft heads North at 320 km/h rel:
AURORKA [14]

The velocity of the aircraft relative to the ground is 240 km/h North

Explanation:

We can solve this problem by using vector addition. In fact, the velocity of the aircraft relative to the ground is the (vector) sum between the velocity of the aircraft relative to the air and the velocity of the air relative to the ground.

Mathematically:

v' = v + v_a

where

v' is the velocity of the aircraft relative to the ground

v is the velocity of the aircraft relative to the air

v_a is the velocity of the air relative to the ground.

Taking north as positive direction, we have:

v = +320 km/h

v_a = -80 km/h (since the air is moving from North)

Therefore, we find

v'=+320 + (-80) = +240 km/h (north)

Learn more about vector addition:

brainly.com/question/4945130

brainly.com/question/5892298

#LearnwithBrainly

7 0
2 years ago
The tip of the fan blade is 0.61 m from the center of the fan. The fan turns at a constant speed and completes 2 rotation every
Goryan [66]

Answer:

What is the centripetal acceleration of the tip of the fan blade?

6.0 m/s2

48 m/s2

53 m/s2

96 m/s2

Answer is 96

Explanation:

5 0
2 years ago
A block of mass 2.00 kg is initially at rest at x=0 on a slippery horizontal surface for which there is no friction. Starting at
Allisa [31]

Answer:

   x = 1,185 m ,     t = 4/3 s ,  F = - 4 N

Explanation:

For this exercise we use Newton's second law

         F = m a = m dv /dt

        β - α t = m dv / dt

        dv = (β – α t) dt

     

We integrate

        v = β t - ½ α t²

We evaluate between the lower limits v = v₀ for t = 0 and the upper limit v = v for t = t

       v-v₀ = β t - ½ α t²

the farthest point of the body is when v = v₀ = 0

  0 = β t - ½ α t²

  t = 2 β / α

  t = 2 4/6

  t = 4/3 s

Let's find the distance at this time

   v = dx / dt

   dx / dt = v₀ + β t - ½ α t2

   dx = (v₀ + β t - ½ α t2) dt

We integrate

   x = v₀ t + ½ β t - ½ 1/3 α t³

   x = v₀ 4/3 + ½ 4 (4/3)² - 1/6 6 (4/3)³

The body comes out of rest

    x = 3.5556 - 2.37

    x = 1,185 m

The value of force is

    F = β - α t

    F = 4 - 6 4/3

   F = - 4 N

8 0
2 years ago
If the radius of the sun is 7.001×105 km, what is the average density of the sun in units of grams per cubic centimeter? The vol
xenn [34]

Answer:

Average density of Sun is 1.3927 \frac{g}{cm}.

Given:

Radius of Sun = 7.001 ×10^{5} km = 7.001 ×10^{10} cm

Mass of Sun = 2 × 10^{30} kg = 2 × 10^{33} g

To find:

Average density of Sun = ?

Formula used:

Density of Sun = \frac{Mass of Sun}{Volume of Sun}

Solution:

Density of Sun is given by,

Density of Sun = \frac{Mass of Sun}{Volume of Sun}

Volume of Sun = \frac{4}{3} \pi r^{3}

Volume of Sun = \frac{4}{3} \times 3.14 \times [7.001 \times 10^{10}]^{3}

Volume of Sun = 1.436 × 10^{33} cm^{3}

Density of Sun = \frac{ 2\times 10^{33} }{1.436 \times 10^{33} }

Density of Sun = 1.3927 \frac{g}{cm}

Thus, Average density of Sun is 1.3927 \frac{g}{cm}.

4 0
2 years ago
Two thin lenses with a focal length of magnitude 12.0cm, the first diverging and the second converging, are located 9.00cm apart
attashe74 [19]

Answer:

Explanation:

b ) First is concave lens with focal length f₁ = - 12 cm .

object distance u = - 20 cm .

Lens formula

1 / v - 1 / u = 1 / f

1 / v + 1 / 20 = -1 / 12

1 / v =  - 1 / 20  -1 / 12

= - .05 - .08333

= - .13333

v = - 1 / .13333

= - 7.5 cm

first image is formed before the first lens on the side of object.

This will become object for second lens

distance from second lens = 7.5 + 9 = 16.5 cm

c )

For second lens

object distance u = - 16.5 cm

focal length f₂ = + 12 cm ( lens is convex )

image distance = v

lens formula ,

1 / v - 1 / u = 1 / f₂

1 / v + 1 / 16.5 = 1 / 12

1 / v =   1 / 12 -  1 / 16.5

= .08333- .0606

= .02273

v = 1 /  .02273

= 44 cm ( approx )

It will be formed on the other side of convex lens

distance from first lens

= 44 + 9 = 53 cm .

magnification by first lens = v / u

= -7.5 / -20 = .375 .

magnification by second lens = v / u

= 44 / - 16.5

= - 2.67

d )

total magnification

= .375 x - 2.67

= - 1.00125

height of final image

= 2.50 mm x 1.00125

= 2.503mm

e )

The final image will be inverted with respect to object  because total magnification is negative .

6 0
2 years ago
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