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IceJOKER [234]
3 years ago
5

I need help solving this,

Physics
1 answer:
Readme [11.4K]3 years ago
8 0

Answer:

8n

Explanation:

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Find the weight required to stretch a steel road by 2 mm, if the length of the road is 2 m and diameter 4 mm. Y= 200 GPa and g -
svetlana [45]

Answer:

Option D is the correct answer.

Explanation:

We equation for elongation

   \Delta L=\frac{PL}{AE}      

Here we need to find weight required,

We need to stretch a steel road by 2 mm, that is ΔL = 2mm = 0.002 m

A=\frac{\pi d^2}{4}=\frac{\pi ({4\times 10^{-3}})^2}{4}=1.26\times 10^{-5}m^2

E = 200GPa = 2 x 10¹¹ N/m²

L=2m

Substituting

0.002=\frac{P\times 2}{1.26\times 10^{-5}\times 2\times 10^{11}}\\\\P=2520N=256.4kg

Option D is the correct answer.

6 0
3 years ago
An electronics technician wishes to construct a parallelplate capacitor using rutile (k=100) as the dielectric. The area of the
My name is Ann [436]

Given k=100, Area if plate A = 1.00 cm2

    = 1 x (10-2)2 m^{2}

Thickness of rutile ds = t = 1.00 mm

      = 1 x 10-3 cm

Without any dielectric medium capacitance of a parallel plate capacitor is c = oAd

When a dielectric is placed between capacitors plates then capacitance c1 = KoAd

Here d = t = 1 x 10-1 cm, o = 8.854 x 10-12 c^{2} m^{-2}  N

Now c1 = 100 X 8.859 X 10^{-12}  X 1 X (10-2)^{21}  X 10^{-2}

= 8.859 X 10-1910-3= 8.85 X 10-11 F= 8.85 X 10-12 F

 C1  = 88.5 X 10-12 F

 C1  = 88.5  PF

Correct Option: A

What purposes serve parallel plate capacitors?

The following are some uses for parallel plate capacitors:

  • This kind of capacitor is utilized in rechargeable energy systems such as batteries.
  • These capacitors are used in systems for dynamic digital memory.
  • Such capacitors are used in pulsed laser and radar circuits.

To learn more about Parallel plate capacitors, visit:

brainly.com/question/12733413

#SPJ4

5 0
2 years ago
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