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stiks02 [169]
2 years ago
10

An algorithm takes 5 seconds for input size 200. how long will it take for input size 1000 if the running time is linear (assume

low-order terms are negligible).
Physics
1 answer:
PSYCHO15rus [73]2 years ago
5 0
It would take 25 seconds
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A 70mm long blockhas cross-section of 50mm by 10mm the block is subjected to forces 60KN (tension) on the 50mm by 10mm face and
sammy [17]

Answer:

970 kN

Explanation:

The length of the block = 70 mm

The cross section of the block = 50 mm by 10 mm

The tension force applies to the 50 mm by 10 mm face, F₁ = 60 kN

The compression force applied to the 70 mm by 10 mm face, F₂ = 110 kN

By volumetric stress, we have that for there to be no change in volume, the total pressure applied by the given applied forces should be equal to the pressure removed by the added applied force

The pressure due to the force F₁ = 60 kN/(50 mm × 10 mm) = 120 MPa

The pressure due to the force F₂ = 110 kN/(70 mm × 10 mm) = 157.142857 MPa

The total pressure applied to the block, P = 120 MPa + 157.142857 MPa = 277.142857 MPa

The required force, F₃ = 277.142857 MPa × (70 mm × 50 mm) = 970 kN

7 0
2 years ago
A speed skater moving across frictionless ice at 8.4 m/s hits a 5.7 m -wide patch of rough ice. She slows steadily, then continu
malfutka [58]

Answer:

Acceleration, a=-2.48\ m/s^2

Explanation:

Initial speed of the skater, u = 8.4 m/s

Final speed of the skater, v = 6.5 m/s

It hits a 5.7 m wide patch of rough ice, s = 5.7 m

We need to find the acceleration on the rough ice. The third equation of motion gives the relationship between the speed and the distance covered. Mathematically, it is given by :

v^2-u^2=2as

a=\dfrac{v^2-u^2}{2s}

a=\dfrac{(6.5)^2-(8.4)^2}{2\times 5.7}

a=-2.48\ m/s^2

So, the acceleration on the rough ice -2.48\ m/s^2 and negative sign shows deceleration.

8 0
3 years ago
The closer together a magnets field lines are,
elena55 [62]
The stronger they will be
4 0
3 years ago
You toss a walnut at a speed of 15.0 m/s at an angle of 50.0∘ above the horizontal. The launch point is on the roof of a buildin
kaheart [24]

Answer:

a.3.51s

b.33.8m

c. 9.64,-22.9

Explanation:

8 0
2 years ago
You apply a very small force, say 0.001 newtons, to a very large truck, with a mass of 2000 kilograms. What can you say for sure
Fofino [41]

Answer:

it will stay still. unless its in space.

Explanation:

5 0
2 years ago
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