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2 years ago

write an interface downloadable that has a method "geturl" that returns the url of a downloadable object

Engineering

1 answer:

Scrat [10]2 years ago

3 0

Answer:

I want to believe the program is to be written in java and i hope your question is complete. The code is in the explanation section below

Explanation:

import java.util.Date;

public interface Downloadable {

//abstract methods

public String getUrl();

public Date getLastDownloadDate();

}

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While there are many ways to solve this problem, one strategy is to calculate the volume of any metal's unit cell given its theo

IgorC [24]

Answer:

$V_{c}=1.396x10^{-28} m^{3} /unit.cell$

Explanation:

z = number of atoms

M = Molar mass of zirconium

N = Avogadro’s number

Vc = volume of zirconium unit cell

d = density

$z=12x\frac{1}{6}+2x\frac{1}{2}+3=6$

z = 6 atoms per unit cell

M = 91.224 g/mol

N = $6.023x10^{23} atoms/mol$

d = $6.51g/cm^{3}$

$V_{c}=\frac{zxM}{dxN}$

$V_{c}=\frac{6x(91.224g/mol)}{(6.51g/cm^{3}) x(6.023x10^{23}atoms/mol) }$

$V_{c}=1.396x10^{-22} cm^{3} /unit.cell$

$V_{c}=1.396x10^{-28} m^{3} /unit.cell$

8 0

2 years ago

Calculate the reactions at 4 ends (supports) of this bookshelf. Assume that the weight of each book is approximately 1 lb. The w

hoa [83]

Answer:

R = 4.75 lb (↑)

Explanation:

Number of books = n = 19

Weight of each book = W = 1 lb

Length of the bookshelf = L = 40 inches

We can get the value of the distributed load as follows

q = n*W/L = 19*1 lb/ 40 inches = 0.475 lb/in

then the reactions at 4 ends (supports) of the bookshelf are

R = (q/2)/2 = 4.75 lb

We can see the bookshelf in the pic.

8 0

2 years ago

A hemispherical shell with an external diameter of 500 mm and a thickness of 20 mm is going to be made by casting, located entir

Olenka [21]

Solution :

Given :

External diameter of the hemispherical shell, D = 500 mm

Thickness, t = 20 mm

Internal diameter, d = D - 2t

= 500 - 2(20)

= 460 mm

So, internal radius, r = 230 mm

= 0.23 m

Density of molten metal, ρ = $7.2 \ g/cm^3$

= $7200 \ kg/m^3$

The height of pouring cavity above parting surface is h = 300 mm

= 0.3 m

So, the metallostatic thrust on the upper mold at the end of casting is :

$F=\rho g A h$

Area, A $=2 \pi r^2$

$=2 \pi (0.23)^2$

$=0.3324 \ m^2$

$F=\rho g A h$

$=7200 \times 9.81 \times 0.3324 \times 0.3$

= 7043.42 N

3 0

2 years ago

In __________, the air bags system checks itself for problems each time the ignition is turned on

zlopas [31]

Answer:

===========

Automobiles

===========

8 0

1 year ago

The product second moment of area Ixy is found by multiplying Ix and Iy. a)True b)- False

Kamila [148]

Answer:

(b)False

Explanation:

$I_{xy}$ defined as

$I_{xy}$ = $\int \left (x\cdot y\right )dA$

Where x is the distance from centroidal x-axis

y is the distance from centroidal y-axis

dA is the elemental area.

The product of x and y can be positive or negative ,so the value of $I_{xy}$ can be positive as well as negative .

So from the above expressions we can say that the product of $I_{x},I_y$ is different from $I_{xy}$ .

7 0

2 years ago