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The atomic number of beryllium (Be) is 4, and the atomic number of barium (Ba) is 56. Which comparison is best supported by this

Vesnalui [34]

The atomic number of beryllium (Be) is 4, and the atomic number of barium (Ba) is 56. the <span>comparison is best supported by this information is that beryllium has a lower atomic radius than Barium</span>

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3 years ago

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devlian [24]

Answer:

cell pohnes convert sound waves into radio waves

electromagnetic waves used in cellphone communications are called: radio waves

To send out a radio signal far and wide.. it is called: broadcast

Explanation:

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3 years ago

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identified a problem in his/her community related to Earth Science. complete research to identify solutions to the problem, deve

son4ous [18]

Answer:

One of the best ways to gain insights in a Design Thinking process is to carry out some form of prototyping. This method involves producing an early, inexpensive, and scaled down version of the product in order to reveal any problems with the current design. Prototyping offers designers the opportunity to bring their ideas to life, test the practicability of the current design, and to potentially investigate how a sample of users think and feel about a product.

Prototypes are often used in the final, testing phase in a Design Thinking process in order to determine how users behave with the prototype, to reveal new solutions to problems, or to find out whether or not the implemented solutions have been successful. The results generated from these tests are then used to redefine one or more of the problems established in the earlier phases of the project, and to build a more robust understanding of the problems users may face when interacting with the product in the intended environment.

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7 0

3 years ago

Asexual reproduction involves only one parent. The offspring of this type of reproduction have - F DNA identical to the DNA of t

Sedbober [7]

Answer:

DNA identical to the DNA of the parent

Explanation:

5 0

3 years ago

Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface.

elena-s [515]

Answer:

$g=13.42\frac{m}{s^2}$

Explanation:

1) Notation and info given

$\rho_{center}=13000 \frac{kg}{m^3}$ represent the density at the center of the planet

$\rho_{surface}=2100 \frac{kg}{m^3}$ represent the densisty at the surface of the planet

r represent the radius

$r_{earth}=6.371x10^{6}m$ represent the radius of the Earth

2) Solution to the problem

So we can use a model to describe the density as function of the radius

$r=0, \rho(0)=\rho_{center}=13000 \frac{kg}{m^3}$

$r=6.371x10^{6}m, \rho(6.371x10^{6}m)=\rho_{surface}=2100 \frac{kg}{m^3}$

So we can create a linear model in the for y=b+mx, where the intercept b= $\rho_{center}=13000 \frac{kg}{m^3}$ and the slope would be given by $m=\frac{y_2-y_1}{x_2-x_1}=\frac{\rho_{surface}-\rho_{center}}{r_{earth}-0}$

So then our linear model would be

$\rho (r)=\rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r$

Since the goal for the problem is find the gravitational acceleration we need to begin finding the total mass of the planet, and for this we can use a finite element and spherical coordinates. The volume for the differential element would be $dV=r^2 sin\theta d\phi d\theta dr$ .

And the total mass would be given by the following integral

$M=\int \rho (r) dV$

Replacing dV we have the following result:

$M=\int_{0}^{2\pi}d\phi \int_{0}^{\pi}sin\theta d\theta \int_{0}^{r_{earth}}(r^2 \rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r)$

We can solve the integrals one by one and the final result would be the following

$M=4\pi(\frac{r^3_{earth}\rho_{center}}{3}+\frac{r^4_{earth}}{4} \frac{\rho_{surface}-\rho_{center}}{r_{earth}})$

Simplyfind this last expression we have:

$M=\frac{4\pi\rho_{center}r^3_{earth}}{3}+\pi r^3_{earth}(\rho_{surface}-\rho_{center})$

$M=\pi r^3_{earth}(\frac{4}{3}\rho_{center}+\rho_{surface}-\rho_{center})$

$M=\pi r^3_{earth}[\rho_{surface}+\frac{1}{3}\rho_{center}]$

And replacing the values we got:

$M=\pi (6.371x10^{6}m)^2(\frac{1}{3}13000 \frac{kg}{m^3}+2100 \frac{kg}{m^3})=8.204x10^{24}kg$

And now that for any shape the gravitational acceleration is given by:

$g=\frac{MG}{r^2_{earth}}=\frac{(6.67408x10^{-11}\frac{m^3}{kgs^2})*8.204x10^{24}kg}{(6371000m)^2}=13.48\frac{m}{s^2}$

4 0

2 years ago