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3 years ago

5

I have an algorithm that runs in O(N ), where N is the size of the problem. For N = 100, the time for the algorithm to run is 1

minute. How long does the algorithm take for N=1000? A) Same time B) 10 minutes C) 100 minutes D) 1000 minutes

1 answer:

hodyreva [135]3 years ago

4 0

Answer:

B) 10 minutes

Explanation:

When the size of the problem is 100 then it takes 1 minute to solve

$100\ s=1\ min$

So, when the size of the problem is 1 then it takes the following minutes

$1\ s=\frac{1}{100}\ min$

For when the size of the problem is 1000 we have

$1000\ s=\frac{1000}{100}\ min=10\ min$

So, it will take 10 minutes to solve the algorithm of size 1000

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A 594 Ω resistor, an uncharged 1.3 μF capacitor, and a 6.53 V emf are connected in series. What is the current in milliamps afte

ivanzaharov [21]

Answer:

6.88 mA

Explanation:

Given:

Resistance, R = 594 Ω

Capacitance = 1.3 μF

emf, V = 6.53 V

Time, t = 1 time constant

Now,

The initial current, I₀ = $\frac{\textup{V}}{\textup{R}}$

or

I₀ = $\frac{\textup{6.53}}{\textup{594}}$

or

I₀ = 0.0109 A

also,

I = $I_0[1-e^{-\frac{t}{\tau}}]$

here,

τ = time constant

e = 2.717

on substituting the respective values, we get

I = $0.0109[1-e^{-\frac{\tau}{\tau}}]$

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or

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or

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3 years ago

17. (16 pts) You set up a two-slit experiment with a laser that produces light with a

Natali [406]

when the two waves interfere with eachother to make a dark spot the periodic difference of the two waves is π . the wave length for 2π is 600nm

. ie. for π difference it is 300nm

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3 years ago

You are driving your car along a country road at a speed of 27.0 m/s. as you come over the crest of a hill, you notice a farm tr

Bezzdna [24]

speed of the car = 27 m/s

speed of truck ahead = 10 m/s

relative speed of car with respect to truck

$v_r = 27 - 10 = 17 m/s$

relative deceleration of car

$a_r = -7 m/s^2$

now the distance before they stop with respect to each other is given by

$v_f^2 - v_i^2 = 2 a d$

$0 - 17^2 = 2 *(-7)*d$

$d = 20.6 m$

so it will come at the same speed of truck after 20.6 m distance and hence it will not hit the truck as the distance of the truck is 25 m from car

Part b)

Distance traveled by car before it stops is given by

$v_f^2 - v_i^2 = 2 a s$

$0^2 - 27^2 = 2 * (-7)* s$

$s = 52.1 m$

so it will stop after it will cover total 52.1 m distance

Part c)

time taken by the car to stop

$v_f - v_i = at$

$0 - 27 = (-7) * t$

$t = 3.86 s$

now the distance covered by truck in same time

$d = 3.86 * 10 = 38.6 m$

now after the car will stop its distance from the truck is

$D = 25 + 38.6 - 52.1 = 11.5 m$

<em>so the distance between them is 11.5 m</em>

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3 years ago

If your apparatus were to be dropped from a mile above the ground, describe the forces acting upon your apparatus as it fell. Ho

kvv77 [185]

Answer:

An accelerometer is a tool that measures proper acceleration.[1] Proper acceleration is the acceleration (the rate of change of velocity) of a body in its own instantaneous rest frame;[2] this is different from coordinate acceleration, which is acceleration in a fixed coordinate system. For example, an accelerometer at rest on the surface of the Earth will measure an acceleration due to Earth's gravity, straight upwards[3] (by definition) of g ≈ 9.81 m/s2. By contrast, accelerometers in free fall (falling toward the center of the Earth at a rate of about 9.81 m/s2) will measure zero.

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When two or more accelerometers are coordinated with one another, they can measure differences in proper acceleration, particularly gravity, over their separation in space—that is, the gradient of the gravitational field. Gravity gradiometry is useful because absolute gravity is a weak effect and depends on the local density of the Earth, which is quite variable.

Single- and multi-axis accelerometers can detect both the magnitude and the direction of the proper acceleration, as a vector quantity, and can be used to sense orientation (because the direction of weight changes), coordinate acceleration, vibration, shock, and falling in a resistive medium (a case in which the proper acceleration changes, increasing from zero). Micromachined microelectromechanical systems (MEMS) accelerometers are increasingly present in portable electronic devices and video-game controllers, to detect changes in the positions of these devices.

Explanation:

hope this helps !!!!

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2 years ago

You attach a meter stick to an oak tree, such that the top of the meter stick is 2.27 meters above the ground. later, an acorn f

Alexandra [31]

The acorn was at a height of <u>4.15 m</u> from the ground before it drops.

The acorn takes a time t to fall through a distance h₁, which is the length of the scale. When the acorn reaches the top of the scale, its velocity is u.

Calculate the speed of the acorn at the top of the scale, using the equation of motion,

$s=ut+ \frac{1}{2} at^2$

Since the acorn falls freely under gravity, its acceleration is equal to the acceleration due to gravity g.

Substitute 2.27 m for s (=h₁), 0.301 s for t and 9.8 m/s² for a (=g).

$s=ut+ \frac{1}{2} at^2\\ (2.27 m)=u(0.301s)+\frac{1}{2}(9.8m/s^2)(0.301s)^2\\ u=\frac{1.8261m}{0.301s} =6.067m/s$

If the acorn starts from rest and reaches a speed of 6.067 m/s at the top of the scale, it would have fallen a distance h₂ to achieve this speed.

Use the equation of motion,

$v^2=u^2+2as$

Substitute 6.067 m/s for v, 0 m/s for u, 9.8 m/s² for a (=g) and h₂ for s.

$v^2=u^2+2as\\ (6.067m/s)^2=(0m/s)^2+2(9.8m/s^2)h_2\\ h_2=\frac{(6.067m/s)^2}{2(9.8m/s^2)} =1.878 m$

The height h above the ground at which the acorn was is given by,

$h=h_1+h_2=(2.27 m)+(1.878 m)=4.148 m$

The acorn was at a height <u>4.15m</u> from the ground before dropping down.

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2 years ago