External depreciation may be defined as a loss in value caused by an undesirable or hazardous influence offsite.
<h3>What is depreciation?</h3>
Depreciation may be defined as a situation when the financial value of an acquisition declines over time due to exploitation, fray, and incision, or obsolescence.
External depreciation may also be referred to as "economic obsolescence". It causes a negative influence on the financial value gradually.
Therefore, it is well described above.
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Answer:
0 m/s , 3 m/s , 2 m/s^2
Explanation:
Given : s(t) = ( t^2 - 6t + 5)
v(t) = ds / dt = 2t - 6
s(0) = 5 m
s(6) = (6)^2 - 6*6 + 5 = 5 m
Vavg = ( s(6) - s(0) ) / 2 = 0 m\s
Find the turning point of particle:
ds/dt = 0 = 2t - 6
t = 3 sec
s(3) = 3^2 -6*3 + 5 = - 4
Total distance = 5 - (-4) + (5 - (-4)) = 18 m
Total time = 6s
Average speed = Total distance / Total time = 18 / 6 = 3 m/s
Taking derivative of v(t) to obtain a(t)
a (t) = dv(t) / dt = 2 m/s^2
Answer:
(a) 0 kJ
(b) 9.81 kJ
(c) 31.32 m/s
Explanation:
(a)
From the law of conservation of energy, energy can only be transformed from one state to another. At a height of 50 m, all the kinetic energy is converted to potential energy hence KE=0
(b)
Potential energy, PE=mgh where m is the mass, g is acceleration due to gravity and h is the height
Substituting 50 m for h and 20 Kg for m, taking g as 9.81 then
PE=20*9.81*50=9810 J=9.81 kJ
(c)
Relating the equation of potential energy to the equation of kinetic energy, which is 
where v is the velocity of the mass
Substituting 50 m for h and taking g as 9.81 then
Answer:
1. Low power hand tools/small
2. Light to medium industrial tools
3. Large industrial tools
There are definitely a lot more categories than three, but this is what I have.
