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Hitman42 [59]
3 years ago
12

As a top-level executive at your own company, you are worried that your employees may steal confidential data too easily by down

loading and taking home data onto thumb drives. What is the best way to prevent this from happening? Install a technical control to prevent the use of thumb drives.
Engineering
1 answer:
DENIUS [597]3 years ago
6 0

Answer:

Create and enforce a written company policy against the use of thumb drives, and install a technical controls on the computers that will prevent the use of thumb drives.

Explanation:

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A lightweight electric car is powered by ten 12-V batteries. At a speed of 100 km/h, the average frictional force is 1500 N. Wha
Aleksandr [31]

Answer:41.67kW

Explanation: a) power = force * velocity

Convert km/hr to m/s=1000m/(60×60s)

=1000/3600

P=F×vel

Power =1500×100× 1000/3600

=1500×100×5/18

=41,666.67W

=41.67kW

6 0
3 years ago
Hi I don't know of yall remeber me, but I'm Jadin aka J. I am looking for my friend group, that I have missed but can't find cau
kirill [66]

Answer:

The young lady was his daughter.The shoemaker was frightened when he saw that she wants to sit near him and took his knife to frighten her and leave him alone to do his work

Explanation:

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7 0
2 years ago
A cylinder with a frictionless piston contains 0.05 m3 of air at 60kPa. The linear spring holding the piston is in tension. The
AleksAgata [21]

Answer:

18 kJ

Explanation:

Given:

Initial volume of air = 0.05 m³

Initial pressure = 60 kPa

Final volume = 0.2 m³

Final pressure = 180 kPa

Now,

the Work done by air will be calculated as:

Work Done = Average pressure × Change in volume

thus,

Average pressure = \frac{60+180}{2}  = 120 kPa

and,

Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³

Therefore,

the work done = 120 × 0.15 = 18 kJ

4 0
3 years ago
What are the optical properties of steel
dezoksy [38]

Answer:

A selective surface with large absorption for solar radiation and high reflectance for thermal infrared radiation was produced by use of surface oxidation of stainless steel. The surfaces were studied for use with concentrated light in a solar power plant at temperatures of 400°C and higher.

In order to investigate the relation between surface treatment and optical properties, stainless steels (AISI 304 and 430) which were submitted to different chemical and mechanical surface treatments, were used. To increase the spectral selectivity, these surfaces were treated in air and in vacuum at different temperatures and times. The optical properties of these films were investigated. Visual and infrared spectral absorptances were measured at room temperature. The thermal hemispherical emittance and absorptance were obtained by a calorimetric method at 200°C. It was noticed that these chemically and mechanically treated stainless steel surfaces have good spectral properties without further oxidations. This is very important for high temperature uses. The best values are found for samples 7 and 8 under vacuum and air. These two samples with mechanically ground surfaces retained their selectivity and specularity after several hours oxidation. One can conclude that the surface ground treatment confers good selectivity on the steel surfaces for use in concentrating solar collectors with a working temperature of 500°C.

Sample surfaces were subjected to long temperature ageing tests in order to gain some idea of the thermal stability of the surfaces. The results promise better-performing surface and the production of durable selective finishes at, possibly, lower cost than competing processes.

Explanation:

3 0
3 years ago
A car accelerates from rest with an acceleration of 5 m/s^2. The acceleration decreases linearly with time to zero in 15 s, afte
Tpy6a [65]

Answer: At time 18.33 seconds it will have moved 500 meters.

Explanation:

Since the acceleration of the car is a linear function of time it can be written as a function of time as

a(t)=5(1-\frac{t}{15})

a=\frac{d^{2}x}{dt^{2}}\\\\\therefore \frac{d^{2}x}{dt^{2}}=5(1-\frac{t}{15})

Integrating both sides we get

\int \frac{d^{2}x}{dt^{2}}dt=\int 5(1-\frac{t}{15})dt\\\\\frac{dx}{dt}=v=5t-\frac{5t^{2}}{30}+c

Now since car starts from rest thus at time t = 0 ; v=0 thus c=0

again integrating with respect to time we get

\int \frac{dx}{dt}dt=\int (5t-\frac{5t^{2}}{30})dt\\\\x(t)=\frac{5t^{2}}{2}-\frac{5t^{3}}{90}+D

Now let us assume that car starts from origin thus D=0

thus in the first 15 seconds it covers a distance of

x(15)=2.5\times 15^{2}-\farc{15^{3}}{18}=375m

Thus the remaining 125 meters will be covered with a constant speed of

v(15)=5\times 15-\frac{15^{2}}{6}=37.5m/s

in time equalling t_{2}=\frac{125}{37.5}=3.33seconds

Thus the total time it requires equals 15+3.33 seconds

t=18.33 seconds

3 0
3 years ago
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