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Natali5045456 [20]
2 years ago
5

An algorithm takes 1.0 ms for input size 100. how long will it take for input size 10000 if the running time is logarithmic (ass

ume low-order terms are negligible).
Physics
1 answer:
xxMikexx [17]2 years ago
5 0

It will take 10000ms

For N = 100 input size

∴ N^2 = 10000 steps and the running time is 1.0 ms.

For N = 10000 input size

∴ N^2 = 100000000 steps are required and running time is

(100000000*1.0)/10000 = 10000 ms is the required running time for execution.

An algorithm is a procedure used for solving a problem or performing a computation. Algorithms act as an exact list of instructions that conduct specified actions step by step in either hardware- or software-based routines.

Learn, more about Algorithms here:

brainly.com/question/22984934

#SPJ4

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A nonconducting sphere is made of two layers. The innermost section has a radius of 6.0 cm and a uniform charge density of −5.0C
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The total charge on the sphere is 46.11 x 10⁻³ C

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radius of the innermost section, r = 6.0cm

charge density of the outer layer ,σ = +8.0 C/m³

radius of the outer layer, r =12.0cm

volume of sphere is given as = \frac{4}{3}.\pi  r^3

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volume = \frac{4}{3}\pi  r^3 = \frac{4}{3}\pi  (0.06^3) = 9.05  X 10^{-4} m^3

Enclosed charge, q₁ = −5.0C/m³ X 9.05 x 10⁻⁴ m³

                             = - 4.53 x 10⁻³ C

Charge at the outer surface

Volume =  \frac{4}{3}\pi  [r_2{^3} - r_1{^3}] = \frac{4}{3}\pi  [0.12{^3} - 0.06{^3}] = 0.00633 {m^3}

Enclosed charge, q₂ = +8.0 C/m³ X 0.00633 m³

                            = 50.64 x 10⁻³ C

Total charge on the sphere; Q = q₁ + q₂

                                                   = - 4.53 x 10⁻³ C +  50.64 x 10⁻³ C

                                                   =  46.11 x 10⁻³ C

Therefore, the total charge on the sphere is 46.11 x 10⁻³ C

3 0
3 years ago
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