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madam [21]
3 years ago
8

A cylindrical capacitor is made of two thin-walled concentric cylinders. The inner cylinder has radius 8 mm , and the outer one

a radius 11 mm . The common length of the cylinders is 152 m . What is the potential energy stored in this capacitor when a potential difference 8 V is applied between the inner and outer cylinder? (k =1/4∗π∗ϵ0=8.99×109N⋅m2/C2) A cylindrical capacitor is made of two thin-walled concentric cylinders. The inner cylinder has radius 8 , and the outer one a radius 11 . The common length of the cylinders is 152 . What is the potential energy stored in this capacitor when a potential difference 8 is applied between the inner and outer cylinder? ( =) 8.5×10−7 J 7.5×10−6 J 2.0×10−7 J 1.4×10−7 J 9.6×10−7 J
Physics
1 answer:
yanalaym [24]3 years ago
3 0

Answer:

So the answer is the first one

E = 8.5 x 10 ⁻⁷ J

Explanation:

Given:

r₁ = 8 mm , r₂ = 11 mm , h = 152 m

ε₀ = 8.85 x 10⁻¹² N * m² / C²

C = Q / V

C = 2π * h * ε₀ / ln ( r₂ / r₁ )

C = [ 2π * 152 m * 8.85 x 10⁻¹² N * m² / C² ]  / [ ln ( 11 mm / 8mm ) ]

C = 2.654 x 10⁻⁸ F

C = 26.54 nF

Now to determine the energy Emf

E = ¹/₂ * C * V²

E = ¹/₂ * 26.54 nF * 8²V

E = 8.5 x 10 ⁻⁷ J

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Suppose a woman raises a 65 N object in 2m in 4 seconds.
Novosadov [1.4K]

Answer:

\huge\boxed{\sf P.E = 130\ Joules}

\huge\boxed{\sf P = 32.5\ Watts}

Explanation:

<u>Given Data:</u>

Weight = W = 65 N

Height = h = 2 m

Time = t = 4 secs

<u>Required:</u>

Power = P = ?

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<u>Formula:</u>

P.E. = Wh

P = P.E. / t

<u>Solution:</u>

P.E. = (65)(2)

P.E = 130 Joules

P = P.E. / t

P = 130 / 4

P = 32.5 Watts

\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807 </h3>
8 0
2 years ago
A 1150 kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 7.69 m before contacting the beam,
Natasha_Volkova [10]

Answer:

the average force 11226 N  

Explanation:

Let's analyze the problem we are asked for the average force, during the crash, we can find this from the impulse-momentum equation, but this equation needs the speeds and times of the crash that we could look for by kinematics.

Let's start looking for the stack speeds, it has a free fall, from rest  (Vo=0)

             

           Vf² = Vo² - 2gY

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This is the speed that the battery likes when it touches the beam.  They also give us the distance it travels before stopping, let's calculate the time

         

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             t = Vo / g

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Now let's use the equation that relates the impulse to the amount of movement

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The amount of final movement is zero because the system stops

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                F = - mv / t

                F = - 1150 12.3 / 1.26

                F = -11226 N

This is the average force exerted by the stack on the vean

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