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expeople1 [14]
4 years ago
14

To water his lawn, a homeowner uses two hoses. One connects to the faucet, the other to the end of the first hose to make the ho

se longer. Each hose is three meters long, but one has 20-mm diameter, the other has 40-mm diameter. When he turns on the faucet, the flow of water at the open end of the second hose measures 10 liters/min. How do the drops in pressure from the start to the end of each hose?
Engineering
1 answer:
Shtirlitz [24]4 years ago
8 0

Answer: to be exact you need 28mm of tubing for that

Explanation:

When the election

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An op-amp is connected in an inverting configuration with R1 = 1kW and R2 = 10kW, and a load resistor connected at the output, R
Svetllana [295]

Answer:

View Image

Explanation:

You didn't provide me a picture of the opamp.

I'm gonna assume that this is an ideal opamp, therefore the input impedance can be assumed to be ∞ . This basically implies that...

  1. no current will go in the inverting(-) and noninverting(+) side of the opamp
  2. V₊ = V₋  , so whatever voltage is at the noninverting side will also be the voltage at the inverting side

Since no current is going into the + and - side of the opamp, then

i₁ = i₂

Since V₊ is connected to ground (0V) then V₋ must also be 0V.

V₊ = V₋  = 0

Use whatever method you want to solve for v_out and v_in then divide them. There's so many different ways of solving this circuit.

You didn't give me what the input voltage was so I can't give you the entire answer. I'll just give you the equations needed to plug in your values to get your answers.

8 0
3 years ago
A controller on an electronic arcade game consists of a variable resistor connected across the plates of a 0.227 μF capacitor. T
anyanavicka [17]

Answer:

R min = 28.173 ohm

R max = 1.55 × 10^{4}  ohm

Explanation:

given data

capacitor = 0.227 μF

charged to 5.03 V

potential difference across the plates =  0.833 V

handled effectively = 11.5 μs to 6.33 ms

solution

we know that resistance range of the resistor is express as

V(t) = V_o \times e^{t\RC}    ...........1

so R will be

R = \frac{t}{C\times ln(\frac{V_o}{V})}    ....................2

put here value

so for t min 11.5 μs

R = \frac{11.5}{0.227\times ln(\frac{5.03}{0.833})}

R min = 28.173 ohm

and

for t max 6.33 ms

R max = \frac{6.33}{11.5} \times 28.173  

R max = 1.55 × 10^{4}  ohm

4 0
3 years ago
In plumbing what is a video snake used for
aleksley [76]

Answer:

How to stop toilets  

Explanation:

I think

Hope this helps

7 0
3 years ago
Read 2 more answers
Errors in the output voltage of an actual integrated circuit operational amplifier can be caused by : Select one:
natta225 [31]

Answer:

Option B

Explanation:

An operational amplifier usually has a high open loop gain of around 10^5 which allows a wide range get of feed back levels in order to achieve the desired performance so therefore a low open loop gain reduces the range feed back level thereby reducing the performance which can cause errors in the output voltage.

7 0
3 years ago
Going green means: increasing one's initiatives toward a concern for the environment. increasing one's bottom line, before any o
Tcecarenko [31]

Answer:

Going green means increasing one's initiatives toward a concern for the environment.

Explanation:

Going green involves all the knowledge and practices that can lead to more environmentally friendly and ecologically responsible decisions and lifestyles, which would protect and sustain the natural resources present in the environment for both present and future generations.

8 0
3 years ago
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