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elena55 [62]
3 years ago
15

Compression is generally lost in 3 places

Engineering
1 answer:
natima [27]3 years ago
4 0

Answer:

leak in cylinder head, piston holes, bad valves.

Explanation:

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A rear wheel drive car of mass 1000 kg is accelerating with a constant acceleration without slipping from 0 to 60 m/s in 1 min.
tester [92]

Answer:

500 N

Explanation:

Given;

Mass of the car, M = 1000 kg

initial speed of the car, u = 0 m/s

Final speed of the car, v = 60 m/s

Time, t = 1 min = 60 s

Now,

Force, F is given as:

F = Ma

where,

a is the acceleration

From the Newton's equation of motion, we have

v = u + at

on substituting the values, we get

60 = 0 + a × 60

or

a = 1 m/s²

Thus,

Force = 1000 × 1 = 1000 N

now,

this force will be equal to the friction force provided by the rear wheels

let the friction force on a single rear wheel be 'f'

thus,

2f = 1000 N

or

f = 500 N

5 0
4 years ago
Consider a 1.80-m-tall man standing vertically in water and completely submerged in a pool. Determine the difference between the
defon

Answer:

17.658 kPa

Explanation:

The hydrostatic pressure of a fluid is the weight of a column of that fluid divided by the base of that column.

P = \frac{weight}{base}

Also, the weight of a column is its volume multiplied by it's density and the acceleration of gravity:

weight = \delta * v * g

Meanwhile, the volume of a column is the area of the base multiplied by the height:

V = base * h

Replacing:

P = \frac{\delta * base * h * g}{base}

The base cancels out, so:

P = \delta * h * g

The pressure depends only on the height of the fluid column, the density of the fluid and the gravity.

If you have two point at different heights (or depths in the case of objects submerged in water) each point will have its own column of fluid exerting pressure on it. Since the density of the fluid and the acceleration of gravity are the same for both points (in the case of hydrostatics density is about constant for all points, it is not the case in the atmosphere), we can write:

\Delta P = \rho * g * (h1 - h2)

We do not know at what depth the man of this problem is, but it doesn't matter, because we know the difference in height of the two points of interes (h1 - h2) = 1.8 m. So:

\Delta P = 9.81 \frac{m}{s^{2} } * 1000 \frac{kg}{m^3} * 1.8 m = 17658 Pa = 17.658 kPa

4 0
3 years ago
Why is the drawdown cone for a well completed in a low permeability aquifer narrower and deeper than a drawdown cone for a well
densk [106]

Answer:

c) It takes a greater hydraulic head to drive the groundwater laterally to the well casing in the lower permeability aquifer

Explanation:

The groundwater are contains under the rock and in the open spaces within the rocks and the unconsolidated sediments. Aquifer refers to the underground layers of the permeable sand or rocks that transmits the groundwater below water table which provides a sufficient supply of water to the well. Groundwater is present everywhere where there is porosity in the rocks and it depends on the permeability of the rocks to allow them flow.

A drawdown cone is completed in the lower permeable aquifer deeper and narrower than the high permeable aquifer as it takes more amount hydraulic head or energy to drive groundwater to the well casing which is in the lower permeable aquifer.

5 0
3 years ago
Valves on steam lines are commonly encountered and you should know how they work. For most valves, the change in velocity of the
damaskus [11]

Answer:

The question is not complete. So, I try to find the complete question & complete answer of the question is given in attached file.

Explanation:

5 0
4 years ago
A plate and frame heat exchanger has 15 plates made of stainless steel that are 1 m tall. The plates are 1 mm thick and 0.6 m wi
hodyreva [135]

Answer:

14.506°C

Explanation:

Given data :

flow rate of water been cooled = 0.011 m^3/s

inlet temp = 30°C + 273 = 303 k

cooling medium temperature = 6°C  + 273 = 279 k

flow rate of cooling medium = 0.02 m^3/s

Determine the outlet temperature

we can determine the outlet temperature by applying the relation below

Heat gained by cooling medium = Heat lost by water

= ( Mcp ( To - 6 )  =  Mcp ( 30 - To )

since the properties of water and the cooling medium ( water ) is the same

= 0.02 ( To - 6 ) = 0.011 ( 30 - To )

= 1.82 ( To - 6 ) = 30 - To

hence To ( outlet temperature ) = 14.506°C

6 0
3 years ago
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