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2 years ago

How much does 1 gallon of water weigh in pound given that the density of water is 1gram/ cm3

Engineering

1 answer:

MAXImum [283]2 years ago

6 0

Explanation:

There are 8.35 pounds in a gallon of water. Water weighs 1 gram per cubic centimeter or 1 000 kilogram per cubic meter, i.e. density of water is equal to 1 000 kg/m³; at 25°C (77°F or 298.15K) at standard atmospheric pressure.

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An energy system can be approximated to simply show the interactions with its environment including cold air in and warm air out

Elenna [48]

Answer: The energy system related to your question is missing attached below is the energy system.

answer:

a) Work done = Net heat transfer

Q1 - Q2 + Q + W = 0

b) rate of work input ( W ) = 6.88 kW

Explanation:

Assuming CPair = 1.005 KJ/Kg/K

Write the First law balance around the system and rate of work input to the system

First law balance ( thermodynamics ) :

Work done = Net heat transfer

Q1 - Q2 + Q + W = 0 ---- ( 1 )

rate of work input into the system

W = Q2 - Q1 - Q -------- ( 2 )

where : Q2 = mCp T = 1.65 * 1.005 * 293 = 485.86 Kw

Q2 = mCp T = 1.65 * 1.005 * 308 = 510.74 Kw

Q = 18 Kw

Insert values into equation 2 above

W = 6.88 Kw

5 0

2 years ago

Side milling cutter is an example of ______ milling cutter.

dusya [7]

Answer:

special type

Explanation:

As per the classification of milling cutters. This cutter can handle deep and long open slots in a more comfortable manner, which increase the productivity.

6 0

2 years ago

Set up the following characteristic equations in the form suited to Evanss root-locus method. Give L(s), a(s), and b(s) and the

Sunny_sXe [5.5K]

Answer:

attached below is the detailed solution and answers

Explanation:

Attached below is the detailed solution

C(iii) : versus the parameter C

The parameter C is centered in a nonlinear equation, therefore the standard locus will not apply hence when you use a polynomial solver the roots gotten would be plotted against C

4 0

3 years ago

What is the metal removal rate when a 2 in-diameter hole 3.5 in deep is drilled in 1020 steel at cutting speed of 120 fpm with a

Studentka2010 [4]

Answer:

a) the metal removal rate is 14.4 in³/min

b) the cutting time is 0.98 min

Explanation:

Given the data from the question

first we find the rpm for the spindle of the drilling tool, using the equation

Ns = 12V/πD

V is the cutting speed(120 fpm) and D is the diameter of the hole( 2 in)

so we substitute

Ns = 12 × 120 / π2

Ns = 1440 / 6.2831

Ns = 229.18 rmp

Now we find the metal removal rate using the equation

MRR = (πD²/4) Fr × Ns

Fr is the feed rate( 0.02 ipr ),

so we substitute

MRR = ((π × 2²)/4) × 0.02 × 229.18

MRR = 14.3998 ≈ 14.4 in³/min

Therefore the metal removal rate is 14.4 in³/min

Next we find the allowance for approach of the tip of the drill

A = D/2

A = 2/2

= 1 in

now find the time required to drill the hole

Tm = (L + A) / (Fr × Ns)

Lis the the depth of the hole( 3.5 in)

so we substitute our values

Tm = (3.5 + 1) / (0.02 × 229.18 )

Tm = 4.5 / 4.5836

Tm = 0.98 min

Therefore the cutting time is 0.98 min

8 0

2 years ago

On highways, the far left lane is usually the _____. A. emergency lane B. merge lane C. slowest D. fastest

slamgirl [31]

On highways, the far left lane is usually the fastest moving traffic.

Answer: Option D.

Explanation:

For the most part, the right lane of a freeway is for entering and leaving the traffic stream. It is an arranging path, for use toward the start and end of your interstate run. The center paths are for through traffic, and the left path is for passing. On the off chance that you are not passing somebody, try not to be driving in the left path.

Regular practice and most law on United States expressways is that the left path is saved for passing and quicker moving traffic, and that traffic utilizing the left path must respect traffic wishing to surpass.

7 0

3 years ago

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