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svlad2 [7]
3 years ago
14

Lightning can be studied with a Van de Graaff generator, which consists of a spherical dome on which charge is continuously depo

sited by a moving belt. Charge can be added until the electric field at the surface of the dome becomes equal to the dielectric strength of air. Any more charge leaks off in sparks as shown in the figure below. Assume the dome has a diameter of 34.5 cm and is surrounded by dry air with a "breakdown" electric field of 4.20*10 6 V/m.
(a) What is the maximum potential of the dome? =..kV
(b) What is the maximum charge on the dome?..μC

Physics
2 answers:
MaRussiya [10]3 years ago
7 0

Answer:

a) Maximum potential of the dome = 724.5kV

b) Maximum charge on the dome = 13.9 micro Coulomb

Explanation:

Diameter of the dome = 34.5 cm =34.5/100 = 0.345 m

Radius of the dome = Diameter/2 = 0.345/2 = 0.1725 m

Electric field strength, E = 4.20 * 10⁶V/m

a) Maximum potential of the dome, V = E*r

V =  4.20 * 10⁶ * 0.1725

V = 724500 Volts = 724.5kV

b) Maximum charge on the dome

V = kQ/r

Q = Vr/k

Q = (724500 * 0.1725)/(9*10⁹)

Q = 1.39 * 10⁻⁵C

Q = 13.9 micro Coulomb

Ymorist [56]3 years ago
5 0

Explanation:

Below is an attachment containing the solution

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Mechanical advantage is a measure of the force amplification achieved by using a tool, mechanical device or machine system.

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A particle with negative charge q and mass m = 2.65×10−15 kg is traveling through a region containing a uniform magnetic field B
Norma-Jean [14]

Answer:

q = 2,95 10-6 C

Explanation:

The magnetic force on a particle is described by the equation

      F = q v x B

Where bold indicate vectors

Let's make the vector product

      vxB =\left[\begin{array}{ccc}i&j&k\\-3&4&12\\0&0&-0.13\end{array}\right]

                     

      v x B = 1.20 106 [i ^ (4 0.130) - j ^ (3 0.130)]

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As they give us the force module, let's use Pythagoras' theorem,

     |v xB | =1.20 10⁶ √( 0.52² + 0.39²)

    |v x B| = 1.20 10⁶ 0.65

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Let's replace and calculate

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4 0
3 years ago
A uniform beam is suspended horizontally by two identical vertical springs that are attached between the ceiling and each end of
puteri [66]

The frequency and amplitude of the SHM beam is 0.8 Hz and 0.098 m. The frequency of the SHM wave when gravel falls is 0.8 Hz and the amplitude of subsequent SHM beam is 0.4m.

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Mass of the sack = 175 Kg

Amplitude of the beam = 40 cm = 0.40 m

Frequency of the beam = F = 0.60 cycles/s

The formula for frequency of oscillation =

= f = (1/2π) X √(k/m)

where, k = 2π²F²m

= k = 2 X (3.14)² X 0.6² X (225 + 175)

= k = 5685.37 N/m

Strength of the spring before gravels fall = x =

= x = mg / k

= x = [ (225 + 175 ) X 9.8 ] / 5685.37

= x = 0.689 m

But, the spring is stretched by the distance of x' which is expressed as,

= X = x - x'

= X = 0.689 - 0.40

= X = 0.289 m

Now, since we know that the gravel falls, thus frequency = f =

= f = (1/2π) X √(k/m)

= f= (1/ 2 X 3.14) X √ 5685.37 / 225

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(b) Assuming that the spring is stretched, x = mg/k =

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Thus, the amplitude of the sack = A = 0.3878 - 0.289

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The frequency = f' =

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= f' = 0.8 Hz

(d) New amplitude = A' =

= A' = 0.38 + 0.038   (after calculating the new distance)

= A' = 0.4 m

To know more about Spring:

brainly.com/question/15850235

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Answer:

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