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Leviafan [203]
2 years ago
12

How much work, in Newtons, is required to lift a 20.4-kg (45lb) plate from the ground to a stand that is 1.50 meters up?

Engineering
1 answer:
nataly862011 [7]2 years ago
4 0

Answer:

Explanation:

Work, U, is equal to the force times the distance:

U = F · r

Force needed to lift the weight, is equal to the weight: F = W = m · g

so:

U = m · g · r

   = 20.4kg · 9.81 \frac{N}{kg} · 1.50m

   = 35.316 \frac{N}{m}

   = 35.316 W

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What are the two most important things to remember when at the end of your interview?
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<h3>Answer:</h3><h3><em>1. Ask questions</em></h3><h3><em>2. Thank the interviewer for their time </em></h3><h3>Explanation:</h3>

1<em>. When the interviewer asked if you have any questions at the end of the interview don't say no. You should always say yes your interviewer is expecting you to ask a few good questions before ending the interview. </em>

<h3><em /></h3>

<em>2. Always thank the interviewer for their time and effort to interview you. This would look very good for you and its a nice way to help wrap up the interview. </em>

5 0
3 years ago
A large well-mixed tank of unknown volume, open to the atmosphere initially, contains pure water. The initial height of the solu
trasher [3.6K]

Answer:

The exact time when the sample was taken is = 0.4167337 hr

Explanation:

The diagram of a sketch of the tank is shown on the first uploaded image

Let A denote the  first inlet

Let B denote the second inlet

Let C denote the single outflow from the tank

From the question we are given that the diameter of A is = 1 cm = 0.01 m

                              Area of  A is  = \frac{\pi}{4}(0.01)^{2} m^{2}

                                                    = 7.85 *10^{-5}m^{2}

Velocity of liquid through A = 0.2 m/s

  The rate at which the liquid would flow through the first inlet in terms of volume  = \frac{Volume of Inlet }{time} = Velocity * Area i.e is m^{2} * \frac{m}{s}   = \frac{m^{3}}{s}

             = 0.2 *7.85*10^{-5} \frac{m^{3}}{s}

  The rate at which the liquid would flow through the first inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

                              =  1039.8 * 0.2 * 7.85 *10^{-5} Kg/s

                              = 0.016324 \frac{Kg}{s}

From the question the diameter of B = 2 cm = 0.02 m

                                           Area of B = \frac{\pi}{4} * (0.02)^{2} m^{2} = 3.14 * 10^{-4}m^{2}

                                     Velocity of liquid through B = 0.01 m/s

The rate at which the liquid would flow through the first inlet in terms of volume  = \frac{Volume of Inlet }{time} = Velocity * Area i.e is m^{2} * \frac{m}{s}   = \frac{m^{3}}{s}

             = 3.14*10^{-4} *0.01 \frac{m^{3}}{s}

The rate at which the liquid would flow through the second inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

                              = 1053 * 3.14*10^{-6} \frac{Kg}{s}

                              = 0.00330642 \frac{Kg}{s}

From the question The flow rate in term of volume of the outflow at the time of measurement is given as  = 0.5 L/s

And also from the question the mass of  potassium chloride  at the time of measurement is given as 13 g/L

So The rate at which the liquid would flow through the outflow in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

                              = 13\frac{g}{L} * 0.5 \frac{L}{s}

                              =  \frac{6.5}{1000}\frac{Kg}{s}       Note (1 Kg = 1000 g)

                              = 0.0065 kg/s

Considering potassium chloride

         Let denote the  rate at which liquid flows in terms of mass as   as \frac{dm}{dt} i.e change in mass with respect to time hence

           Input(in terms of mass flow ) - output(in terms of mass flow ) = Accumulation in the Tank(in terms of mass flow )

         

      (0.016324 + 0.00330642) - 0.0065 = \frac{dm}{dt}

          \int\limits {\frac{dm}{dt} } \, dx  =\int\limits {0.01313122} \, dx

      => 0.01313122 t = (m - m_{o})

  From the question  (m - m_{o})  is given as = 19.7 Kg

Hence the time when the sample was taken is given as

               0.01313122 t = 19.7 Kg

      =>  t = 1500.2414 sec

            t = .4167337 hours (1 hour = 3600 seconds)

5 0
3 years ago
Under which of the following conditions is a Type B-1 Fire extinguisher required onboard a motorized vessel?
swat32

Answer:

The correct option is;

D. The vessel has closed living spaces onboard

Explanation:

Type B-1 Fire extinguishers

A fire extinguisher is required by the law to be installed in a boat that hs the following specifications

1) There are closed compartment in the boat that can be used for fuel storage

2) There exist double double bottom that is only partially filled with flotation materials

3) There are closed living spaces in the boat

4) The fuel tank is permanently installed in the boat

5) The engine is inboard.

7 0
3 years ago
Read 2 more answers
Major processing methods for fiberglass composited include which of the following? Mark all that apply) a)- Open Mold b)- Closed
Novay_Z [31]

Answer:

it is f all of the above

Explanation:

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im not positive if im right but i should be right

4 0
3 years ago
2. When it comes to selling their crop, what are 3 options a farmer has when harvesting their grain?
tiny-mole [99]

Answer:

Sell his crop, use his crop as food, and sell his crop

Explanation:

6 0
3 years ago
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