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1 year ago

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Why is thesize saved prior to entering the for loop? 2. what is the running time of removefirsthalf if lst is an arraylist? 3. w

hat is the running time of removefirsthalf if lst is a linkedlist?

1 answer:

Mrac [35]1 year ago

8 0

<span>After entering the loop, it should use the correct list size and the loop will be affected if the remove call changes the size of the list. If lst is an Arraylist the running time of removefirsthalf is O (n^2). So when the beginning is removed the next element will move forward. If lst is a LinkedList which is a dynamic structure the running is O (n) for removefirsthalf</span>

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Calculate the flux of the vector field F⃗ =−6i⃗ +5x2j⃗ −5k⃗ , through the square of side 8 in the plane y=1, centered on the y-a

Tasya [4]

Answer:

The flux is 682.6 Wb.

Explanation:

Given that,

Vector field $F=-6i+5x^2j-5k$

We need to calculate the flux

Using formula of flux

$\phi=\int_{-4}^{4}\int_{-4}^{4}(F\cdot j\ dxdz)$

Put the value into the formula

$\phi=\int_{-4}^{4}\int_{-4}^{4}(-6i+5x^2j-5k)1\ dxdz$

$\phi=\int_{-4}^{4}\int_{-4}^{4}(5x^2)dxdz$

$\phi=2(\dfrac{x^3}{3})_{-4}^{4}\times(z)_{-4}^{4}$

$\phi=682.6\ Wb$

Hence, The flux is 682.6 Wb.

7 0

1 year ago

Two oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters are separated by a dielectric 1

7nadin3 [17]

Answer:

A). σ = 3.823 x $10^{-5}$ $C^{2}$ /N- $m^{2}$

B). $\sigma ^{'}=2.76\times 10^{-5}$ C/ $m^{2}$

C). $U=10.322$ J

Explanation:

A). We know magnitude of charge per unit area for a conducting plate is given by

$\sigma =k.\varepsilon _{0}.E$

where, E is resultant electric field = 1.2 x $10^{6}$ V/m

$\varepsilon _{0}$ is permittivity of free space = 8.85 x $10^{-12}$ $C^{2}$ /N- $m^{2}$

k is dielectric constant = 3.6

∴ $\sigma =k.\varepsilon _{0}.E$

= 3.6 x 8.85 x $10^{-12}$ x 1.2 x $10^{6}$

= 3.823 x $10^{-5}$ $C^{2}$ /N- $m^{2}$

B).Now we know that the magnitude of charge per unit area on the surface of the dielectric plate is given by

$\sigma ^{'}=\sigma\left ( 1-\frac{1}{k} \right )$

$\sigma ^{'}=3.823\times 10^{-5}\left ( 1-\frac{1}{3.6} \right )$

$\sigma ^{'}=2.76\times 10^{-5}$ C/ $m^{2}$

C).

Area of the plate, A = 2.5 $cm^{2}$

= 2.5 x $10^{-4}$ $m^{2}$

diameter of the plate, d = 1.8 mm

= 1800 m

∴ Total energy stored in the capacitor

$U=\frac{1}{2}k\varepsilon _{0}E^{2}Ad$

$U=\frac{1}{2}\times 3.6\times8.85 \times10^{-12}\times\left ( 1.2\times 10^{6} \right ) ^{2}\times 2.5\times 10^{-4}\times 1800$

$U=10.322$ J

4 0

2 years ago

A 50-kg load is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. By what distance will the wire stretch? Young'

lbvjy [14]

Answer:

3.5 cm

Explanation:

mass, m = 50 kg

diameter = 1 mm

radius, r = half of diameter = 0.5 mm = 0.5 x 10^-3 m

L = 11.2 m

Y = 2 x 10^11 Pa

Area of crossection of wire = π r² = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3

= 7.85 x 10^-7 m^2

Let the wire is stretch by ΔL.

The formula for Young's modulus is given by

$Y =\frac{mgL}{A\Delta L}$

$\Delta L =\frac{mgL}{A\times Y}$

ΔL = 0.035 m = 3.5 cm

Thus, the length of the wire stretch by 3.5 cm.

5 0

1 year ago

A teacher sets up a stand carrying a convex lens of focal length 15 cm at 20.5 cm mark on the optical bench. She asks the studen

Brums [2.3K]

We get the clearest image if there is no magnification. When we have no magnification the image and real object have the same size.
If we look at the diagram that I attached we can see that:
$\frac{h_i}{h_0}=\frac{d_i}{d_0}$
Two triangles that I marked are similar and from this we get:
$\frac{h_i}{h_0}=\frac{d_i-f}{f}$
The image and the object must have the same height so we get:
$\frac{h_i}{h_0}=\frac{d_i-f}{f};h_i=h_0\\
1=\frac{d_i-f}{f}\\
d_i=2f$
This tells how far the screen should be from the lens.
The position of the screen on the optical bench is:
$S=20.5cm+2f=20.5+2\cdot 15cm=50.5cm$

8 0

1 year ago

After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of r

mina [271]

Answer:

t = 103.45 n m

Explanation:

given,

refractive index of cornea = 1.38

refractive index of eye drop = 1.45

wavelength of refractive index = 600 nm

refractive index of eye drop is greater than refractive index of cornea and the air.

Formula used in this case

for constructive interference

$2 n t = (m + \dfrac{1}{2})\lambda$

At m = 0 for the minimum thickness, so

$2\times 1.45 \times t = (0 + 0.5)\times 600$

$2.9 \times t =300$

$t =\dfrac{300}{2.9}$

t = 103.45 n m

the minimum thickness of the film of eyedrops t = 103.45 n m

6 0

2 years ago