Answer:
Buffers in electrical systems are amplifiers that prevent input voltage from being affected by whatever curent the load draws
Explanation:
The input and output parts of the circuit are isolated. By having high-impedance(following ohms law, V=IR) very small current is drawn by the amplifier circuit. The output and input voltages are same. However, the output impedance is very low. In this way power losses are minimized and vlotage levels are maintained for the load
They are useful where a measurement of small signal is required in the presence of high voltage.
They are also used in multi-stage filters to isolate one stage from another
Answer:
# -*- coding: utf-8 -*-
# Get N from the command line
import sys
N = int(sys.argv[1])
if N > 0: #N is positive
positve = list(range(N,0,-1))
print(positve)
elif N < 0: #N is negative
negative = list(range(N,0,1))
print(negative)
else:
print("Invalid in input")
Explanation:
First, you need to identify if the number entered is positive, negative or none of them, for that we use one if, one elif and one else statement:
- If the number entered (N) is greater than zero (is positive) we print a list in the range N to 0 in steps of minus one, the zero is not printed because the function range by default omits the last value
- If the previous statement was False and the number entered is smaller than zero (is negative) we print a list in the range N to 0 in steps of one, the zero is not printed because the function range by default omits the last value
- Finally, if the two previous events were false print invalid input
Answer:
The speed is the same at 1.5 m/s while
The work done by the force F is 0.4335 J
Explanation:
Here we have angular acceleration α = v²/r
Force = ma = 2.8 × 1.5²/r₁
and ω₁ = v₁/r₁ = ω₂ = v₁/r₂
The distance moved by the force = 600 - 300 = 300 mm = 0.3 m
If the velocity is constant
The speed is 1.5 m/s while the work done is
2.8 × 1.5²1/(effective radius) ×0.3
r₁ = effective radius
2.8*9.81 = 2.8 × 1.5²/r₁
r₁ = 0.229
The work done by the force = 2.8 × 1.5²*1/r₁ *0.3 = 0.4335 J
Answer:
8*10000+3*1000+1*00+2*10+2
Explanation:
Answer:
Considering the guidelines of this exercise.
The pieces produced per month are 504 000
The productivity ratio is 75%
Explanation:
To understand this answer we need to analyze the problem. First of all, we can only produce 2 batches of production by the press because we require 3 hours to set it up. So if we rest those 6 hours from the 8 of the shift we get 6, leaving 2 for an incomplete bath. So multiplying 2 batches per day of production by press we obtain 40 batches per day. So, considering we work in this factory for 21 days per month well that makes 40 x 21 making 840 then we multiply the batches for the pieces 840 x 600 obtaining 504000 pieces produced per month. To obtain the productivity ratio we need to divide the standard labor hours meaning 6 by the amount of time worked meaning 8. Obtaining 75% efficiency.