I think that the answer would be B or C
Answer:
Personal computers:
Personal computers may be useful and lead to productivity as using a computer an employee familiars with is a good thing. However, the disadvantages in some facilities especially ones dealing with customer and information security can include data theft, unauthorized data sharing, uses of internet connection for personal purposes, as this can slow down internet connection at the facility, distraction at work place etc.
Hard drive:
Due to large amount of data that can be stored on a hard drive, it might not be allowed in some facilities to avoid data theft and unauthorized transfer.
Music players:
This might be restricted to avoid distraction at work. Noice in places such as libraries would cause abnormality and poor service delivery. An employee with loud speaker at work would not only distracts himself but also other staffs and customers.
PSP Game Device and other game devices:
Playing games during working hour may jeopardize the productivity and therefore might be resctrited in some facilities and working places.
Electronic digital notepad:
Carrying a handheld electronic digital notepad to the work place can cause lack of concentration and division of attention on work and other personal activities. These can harm working harmony and business productivity.
Video recorder:
In some facilities, this device might not be allowed due to facility privacy and protection from unwanted navigation.
Explanation:
Answer:
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Explanation:
Answer:
Heat losses by convection, Qconv = 90W
Heat losses by radiation, Qrad = 5.814W
Explanation:
Heat transfer is defined as the transfer of heat from the heat surface to the object that needs to be heated. There are three types which are:
1. Radiation
2. Conduction
3. Convection
Convection is defined as the transfer of heat through the actual movement of the molecules.
Qconv = hA(Temp.final - Temp.surr)
Where h = 6.4KW/m2K
A, area of a square = L2
= (0.25)2
= 0.0625m2
Temp.final = 250°C
Temp.surr = 25°C
Q = 64 * 0.0625 * (250 - 25)
= 90W
Radiation is a heat transfer method that does not rely upon the contact between the initial heat source and the object to be heated, it can be called thermal radiation.
Qrad = E*S*(Temp.final4 - Temp.surr4)
Where E = emissivity of the surface
S = boltzmann constant
= 5.6703 x 10-8 W/m2K4
Qrad = 5.6703 x 10-8 * 0.42 * 0.0625 * ((250)4 - (25)4)
= 5.814 W