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4 years ago

9

14) Modify the method that calculates the sum of the integers between 1 and N. Have the new version match the following recursiv

e definition: The sum of 1 to N is the sum of 1 to (N/2) plus the sum of (N/2 1) to N. Trace your solution using an N of 7.

1 answer:

Setler [38]4 years ago

8 0

Explanation:

Let solve the program using Java programming language

Method: method means group of statements to perform some operation.

let call the method sum.

Parameters: list of variables that are use in the method for declaration.

<u>The code</u>

public int sum (int number)

int answer;

if (number == 1)

answer = number;

else

{

int half = number/2;

int span = number - half;

answer = sum(half) + sum(span) + (half * span);

}

answer result;

}

Firstly we defined the method called sum(), the method takes only one parameter which is number and return the answer(sum of the integers 1 and N).

if the number is equal to 1, so it will return the number and if the number is not equal to 1 it will divide the number by 2 and get the span(span used to shift upper range).And result will add sum of half, sum of span and product of half span.

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An object that is farther from a converging lens than its focal point always has an image that is _____.

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Answer:

$\displaystyle |F_t|=10.9\ \frac{KQ^2}{l^2}$

$\displaystyle \theta =68^o$

Explanation:

Electrostatic Force

It's the force that appears between two electrical charges q1 q2 when they are placed at a certain distance d. The force can be computed by using the Coulomb's law:

$\displaystyle F=\frac{KQ_1Q_2}{d^2}$

We have an arrangement of 4 charges as shown in the image below. We need to calculate the total force exerted on the charge 2Q by the other 3 charges. The free body diagram is also shown in the second image provided. The total force on 2Q is the vectorial sum of F1, F2, and F3. All the forces are repulsive, since all the charges have the same sign. Let's compute each force as follows:

$\displaystyle |F_1|=\frac{KQ(2Q)}{l^2}=\frac{2KQ^2}{l^2}$

$\displaystyle |F_2|=\frac{K(2Q)(4Q)}{l^2}=\frac{8KQ^2}{l^2}$

The distance between 3Q and 2Q is the diagonal of the rectagle of length l:

$\displaystyle |d_3|=\sqrt{l^2+l^2}=\sqrt{2}\ l$

The force F3 is

$\displaystyle |F_3|=\frac{K(3Q)(2Q)}{(\sqrt{2l)}^2}=\frac{3KQ^2}{l^2}$

Each force must be expressed as vectors. F1 is pointed to the right direction, thus its vertical components is zero

$\displaystyle \vec{F_1}=\left \langle |F_1|,0 \right \rangle=\left \langle \frac{2KQ^2}{l^2},0 \right \rangle$

F2 is pointed upwards and its horizontal component is zero

$\displaystyle \vec{F_2}=\left \langle 0,\frac{8KQ^2}{l^2} \right \rangle$

F3 has two components because it forms an angle of 45° respect to the horizontal, thus

$\displaystyle \vec{F_3}=\left \langle \frac{3KQ^2}{l^2}\ cos45^o,\frac{3KQ2}{l^2} sin45^o\right \rangle$

$\displaystyle \vec{F_3}=\left \langle \frac{3\sqrt{2}KQ^2}{2l^2},\frac{3\sqrt{2}KQ^2}{2l^2}\right \rangle$

Now we compute the total force

$\displaystyle \vec{F_t}=\vec{F_1}+\vec{F_2}+\vec{F_3}$

$\displaystyle \vec{F_t}=\left \langle \frac{2KQ^2}{l^2},0 \right \rangle +\left \langle 0,\frac{8KQ^2}{l^2} \right \rangle + \left \langle \frac{3\sqrt{2}KQ^2}{2l^2},\frac{3\sqrt{2}KQ^2}{2l^2}\right \rangle$

$\displaystyle \vec{F_t}=\left \langle \left(2+\frac{3\sqrt{2}}{2}\right)\frac{KQ^2}{l^2},\left(8+\frac{3\sqrt{2}}{2}\right) \frac{KQ^2}{l^2}\right \rangle$

$\displaystyle F_t=\left \langle 4.121,10.121 \right \rangle \frac{KQ^2}{l^2}$

Now we compute the magnitude

$\boxed{\displaystyle |F_t|=10.9\ \frac{KQ^2}{l^2}}$

The direction of the total force is given by

$\displaystyle tan\theta =\frac{10.121}{4.121}=2.4558$

$\boxed{\displaystyle \theta =68^o}$

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3 years ago

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Answer:

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d = Diameter of wheel = 2 m

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Angular speed is given by

$\omega=\frac{v}{r}\\\Rightarrow \omega=\frac{0.25}{1}\\\Rightarrow \omega=0.25\ rad/s=0.25\times \frac{60}{2\pi}\\\Rightarrow \omega=2.38732\ rpm$

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The angular acceleration of the wheel is 1.22625 rad/s²

Angular displacement is given by

$\theta=\frac{s}{r}\\\Rightarrow \theta=\frac{2.85}{1}\\\Rightarrow \theta=2.85\ rad=2.85\times \frac{360}{2\pi}\\ =163.292\ ^{\circ}$

The angle the disk turned when it has raised the elevator is 163.292°

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3 years ago

A flat screen is located 0.55 m away from a single slit. Light with a wavelength of 544 nm (in vacuum) shines through the slit a

kobusy [5.1K]

Answer:

Explanation:

The diffraction pattern is given as

Sinθ = mλ/ω

Where m=1,2,3,4....

Now, when m=1

Sinθ = λ/ω

Then,

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Given that,

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Substituting this into

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Then,

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3 years ago