Answer:
fluid power
Explanation:
fluids commonly used in fluid power are Oil, Water, Air, CO², and Nitrogen gas, fluid power is commonly confused with hydraulic power, which only uses liquids, fluid power uses either liquids or gases
Given an array of words representing your dictionary, you test words to see if it can be made into another word in dictionary. T
his will be done by removing characters one at a time. Each word represents its own first element of its string chain, so start with a string chain length of 1. Each time you remove a character, increment your string chain by 1. In order to remove a character the resulting word must be in your original dictionary. Your goal is to determine the longest string chain available for a given dictionary. for example, given a dictionary [a, and, ab, bear] the word and could be reduced to an and the word an to a. The single character a can not be reduced to any further as the null string is not in the dictionary. This would be the longest string chain, having a length 3. The word bear can not be reduced at all.
1 answer:
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Answer:
7.615 kW
Explanation:
Solution in pen paper form in the attachment section
Answer:
a) 2,945 mC
b) P(t) = -720*e^(-4t) uW
c) -180 uJ
Explanation:
Given:
i (t) = 6*e^(-2*t)
v (t) = 10*di / dt
Find:
( a) Find the charge delivered to the device between t=0 and t=2 s.
( b) Calculate the power absorbed.
( c) Determine the energy absorbed in 3 s.
Solution:
- The amount of charge Q delivered can be determined by:
dQ = i(t) . dt
- Integrate and evaluate the on the interval:
- The power can be calculated by using v(t) and i(t) as follows:
v(t) = 10* di / dt = 10*d(6*e^(-2*t)) /dt
v(t) = 10*(-12*e^(-2*t)) = -120*e^-2*t mV
P(t) = v(t)*i(t) = (-120*e^-2*t) * 6*e^(-2*t)
P(t) = -720*e^(-4t) uW
- The amount of energy W absorbed can be evaluated using P(t) as follows:
- Integrate and evaluate the on the interval:
Answer:
This is an asynchrnous 3-bit counter. Just note that this design is different and works differently than its synchronous counterpart. It's an easier design than its synchronous counterpart, and is not as reliable because it has delays.
