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natka813 [3]
2 years ago
11

1. Band saw lower wheel does not require a guard * true or false 2. Band saw upper guide should be adjusted to within 1/8" of th

e work piece * true or false 3. Find board & linear ft for 10 pieces of 4" x 4" x 8' *​
Engineering
1 answer:
Blizzard [7]2 years ago
4 0

Answer:

1. Band saw lower wheel does not require a guard * true or false 2. Band saw upper guide should be adjusted to within 1/8" of the work piece * true or false 3. Find board & linear ft for 10 pieces of 4" x 4" x 8' *

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Draw the internal connections of motor generator set​
Akimi4 [234]

Answer:

tyjtgfjhgk vgjyg7igjccxfb  rt5bshe dgrty5rm nry5ghbhjyrdegbtyr45bh4 cnbfgcb xdftjrnn hdftytr s  jhbgfhtyujt ntj

Explanation:

3 0
3 years ago
Given a 12-bit A/D converter operating over a voltage range from ????5 V to 5 V, how much does the input voltage have to change,
Doss [256]

Answer:

2.44 mV

Explanation:

This question has to be one of analog quantization size questions and as such, we use the formula

Q = (V₂ - V₁) / 2^n

Where

n = 12

V₂ = higher voltage, 5 V

V₁ = lower voltage, -5 V

Q = is the change in voltage were looking for

On applying the formula and substitutiting the values we have

Q = (5 - -5) / 2^12

Q = 10 / 4096

Q = 0.00244 V, or we say, 2.44 mV

6 0
3 years ago
A 4140 steel shaft, heat-treated to a minimum yield strength of 100 ksi, has a diameter of 1 7/16 in. The shaft rotates at 600 r
velikii [3]
Answer:










Explanation:



4140-40 I’d pick wood




I hope this helps! :)
4 0
3 years ago
Read 2 more answers
Why can't I fly. I dont know why?
algol13
In order to fly, you must have a device/mechanism that will release hot air, causing it to fly. A jet pack will do the job.
5 0
3 years ago
A waste stabilization pond is used to treat a dilute municipal wastewater before the liquid is discharged into a river. The infl
german

Answer:

BOD concentration at the outflow = 17.83 mg/L

Explanation:

given data

flow rate of Q = 4,000 m³/day

BOD1 concentration of Cin = 25 mg/L

volume of the pond = 20,000 m³

first-order rate constant equal = 0.25/day

to find out

What is the BOD concentration at the outflow of the pond

solution

first we find the detention time that is

detention time t = \frac{volume}{flow rate}

detention time t = \frac{20000}{4000}

detention time = 5 days

so

BOD concentration at the outflow of pond is express as

BOD concentration at the outflow = Cin ( 1 - e^{-kt} )

here k is first-order rate constant and t is detention time and Cin is BOD1 concentration

so

BOD concentration at the outflow = 25 ( 1 - e^{-0.25(5)} )

BOD concentration at the outflow = 17.83 mg/L

8 0
3 years ago
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