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Vesnalui [34]
3 years ago
10

A net charge is placed on a hollow conducting sphere. How does the net charge distribute itself? A net charge is placed on a hol

low conducting sphere. How does the net charge distribute itself? The net charge uniformly distributes itself on the sphere's inner and outer surfaces. The net charge clumps together at some location within the sphere. The net charge uniformly distributes itself on the sphere's outer surface. The net charge uniformly distributes itself throughout the thickness of the conducting sphere. The net charge uniformly distributes itself on the sphere's inner surface.
Physics
1 answer:
makvit [3.9K]3 years ago
5 0

Answer:

The net charge uniformly distributes itself on the sphere's inner and outer surfaces.

Explanation:

In electrostatic condition The charge will be distributed over the surface  of the shell,

The electic field outside the sphere  is like the charge where concentrated in the center of the sphere, inside the sphere the electric field is cero (quivalent to say the potencial is constant, non  electric current flow) If the thickness of the shell is very very thin we just have a charged layer. To keep the above mentioned electrostatic condition the elecrtic field have to be cero in this region  too

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A high diver of mass 51.7 kg steps off a board 10.0 m above the water and falls vertical to the water, starting from rest. If he
zmey [24]

Answer:

851.33 N

Explanation:

Using newton;s equation of motion,

v² = u² + 2gh ......... Equation 1

Where v = final velocity, u = initial velocity, g = acceleration due to gravity, h = height

Given: u = 0 m/s(from rest), h = 10 m g = 9.8 m/s².

Substitute into equation 1

v² = 0² + 2(9.8)(10)

v² = 196

v = √196

v = 14 m/s.

Note: As the Diver touches the water,  u = 14 m/s and v = 0 m/s( stopped)

Using,

d = (v+u)t/2 .............. equation 2

Where d = distance moved by the diver in water before its motion stopped, t = time taken before it comes to rest

Given: v = 0 m/s, u = 14 m/s t = 2.10 s

Substitute into equation 2

d = (0+14)2.1/2

d = 14.7 m.

Finally

work done by the water to stop the diver = potential energy of the diver

F×d = mgh'................Equation 3

Where F = force of the diver in water, d = distance of the diver in the water, m = mass of the diver, g = acceleration due to gravity, h' = height of the diver from the point of fall to the point where he comes to rest

making F the subject of the equation,

F = mgh/d ............ Equation 4

Given: m = 51.7 kg, h = 10 m, g = 9.8 m/s², d = 14.7 m.

Substitute into equation 4

F = 51.7(10+14.7)(9.8)/14.7

F = 851.33 N

Hence the upward force the water exert on her = 851.33 N

8 0
3 years ago
Timbre describes the sound’s _____.<br><br> pitch<br><br> loudness<br><br> tone<br><br> quality
Bumek [7]

Answer:

A) pitch

Explanation:

I did band and also g*ogled it to check :)

If it isn't A, however, it is C

Hope this helps :D

7 0
2 years ago
Read 2 more answers
A small rubber ball is launched by a compressed-air cannon from ground level with an initial speed of 11.8 m/s directly upward.
Fantom [35]

Answer:

7.09683 m

1.20285 s

2.4057 s

11.8 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² (negative up, positive down)

From equation of motion we have

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-11.8^2}{2\times -9.81}\\\Rightarrow s=7.09683\ m

The maximum height above the ground that the ball reaches is 7.09683 m

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-11.8}{-9.81}\\\Rightarrow t=1.20285\ s

Time taken to go up is 1.20285 s it will take the same time to come down so total time taken to reach the ground after it is shot is 1.20285+1.20285 = 2.4057 s

v=u+at\\\Rightarrow v=0+9.81\times 1.20285\\\Rightarrow v=11.8\ m/s

The velocity just before it hits the ground is 11.8 m/s

6 0
2 years ago
Two thin conducting plates, each 56.0 cm on a side, are situated parallel to one another and 7.0 mm apart. If 10^-10 electrons a
dsp73

Answer:

E=576.5V/m

Explanation:

From the question we are told that:

Length l=56.0cm=0.56m

Distance apart d=7.0mm=0.007m

Electron Transferred n=10^{-10}

Therefore

Total Charge

Since Charge on each electron is

e=1.602*10^{-19}

Therefore

T=1.602*10^{-19} *10^{10}

T=1.602*10^{-9}

Generally the equation for Charge density is mathematically given by

\rho=T/A

Where

Area

A=0.56*0.56

A=0.3136

Therefore

\rho=1.602*10^{-9}/0.3136

\rho=5.10*10^{-9}

Generally the equation for Electric Field in the capacitor is mathematically given by

E=\frac{\rho}{e_0}

E=\frac{5.10*10^{-9}}{8.85x10{-12}}

E=576.5V/m

8 0
2 years ago
​Hooke's Law. The distance d when a spring is stretched by a hanging object varies directly as the weight w of the object. If th
nadezda [96]

Answer:

distance when the weight is 8 ​kg is 26.66 cm

Explanation:

given data

distance d2 = 10 cm

weight w2 = 3 ​kg

weight w1 = 8 kg

to find out

distance when the weight is 8 ​kg

solution

we consider here distance d1 when weight is 8 kg

so equation will be

d1/d2 = w1/w2

d/10 = 8/ 3

so d = 8/3 × 10

so d = 26.66

distance when the weight is 8 ​kg is 26.66 cm

7 0
3 years ago
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