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san4es73 [151]
3 years ago
15

Need answers for these please ​

Engineering
1 answer:
Step2247 [10]3 years ago
7 0

Answer:

Following are given the answers one-by-one with explanation:

(a) 8 records

As records are the horizontal rows of data and we have 8 of them.

(b) 5 fields

As fields are vertical columns of data and we have 5 of them.

(c)E3000

As by sorting Makers_names in ascending order the last one would be Rany and by sorting Processor_type we have two option with Rany that are B1500 and E3000. We can see that E3000 will come at the end.

(d) Computer_type

As only two choices (Laptop and PC )  are being used under the column computer_type so it can be amended to contain Boolean data.

(e) format check

This will check that the field should contain one alphabet (capital letter) followed by 4 digits each.

i hope it will help you!

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The Employee

Explanation:

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A satellite is launched 600 km from the surface of the earth, with an initial velocity of 8333.3 m./s, acting parallel to the ta
Vikki [24]

Answer:

eccentrcity of orbit is 0.22

Explanation:

GIVEN DATA:

Initial velocity of satellite = 8333.3 m/s

distance from the sun is 600 km

radius of earth is 6378 km

as satellite is acting parallel to the earth therefore\theta angle = 0

and radial component of given velocity is zero

we haveh = r_o v_r_o = 6378+600 =6.97*10^6 m

h = 6.97*10^6 *8333.3 = 58.08*10^9 m^2/s

we know that

\frac{1}{r} =\frac{GM}{h^2} \times ( i + \epsilon cos\theta)

GM = gr^2 = 9.81*(6.37*10^6)^2 = 398*10^{12} m^3/s

so

\frac{1}{6.97*10^6} =\frac{398*10^{12}}{(58.08*10^9)^2} \times ( i + \epsilon cos0)

solvingt for \epsilon)

\epsilon = 0.22)

therefore eccentrcity of orbit is 0.22

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3 years ago
Why is data driven analytics of interest to companies?
Svetlanka [38]

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Advantages of data analysis

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3 years ago
An electrical current of 700 A flows through a stainlesssteel cable having a diameter of 5 mm and an electricalresistance of 610
KatRina [158]

Answer:

778.4°C

Explanation:

I = 700

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q = I²R

q = 700²(6x10⁻⁴)

= 490000x0.0006

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we calculate the surface temperature

Ts = T∞ + \frac{q}{h\pi Di}

Ts = 30+\frac{294}{25*\frac{22}{7}*\frac{5}{1000}  }

Ts=30+\frac{294}{0.3928} \\

Ts =30+748.4\\Ts = 778.4

The surface temperature is therefore 778.4°C if the cable is bare

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For some metal alloy, the following engineering stresses produce the corresponding engineering plastic strains prior to necking.
kirza4 [7]

Answer:

203.0160

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