35 points an brainiest if correct

Worth 20 points! Please help ASAP!

gulaghasi [49]

Answer:

I am in 6th grade, why are high school things popping up??

Explanation:

8 0

3 years ago

Water enters the pump of a steam power plant as saturated liquid at 20 kPa at a rate of 45 kg/s and exits at 6 MPa. Neglecting t

Natasha2012 [34]

Answer:

$\dot W_{in} = 273.69\,kW$

Explanation:

The pump is modelled after the First Law of Thermodynamics. A reversible process means that fluid does not report any positive change in entropy:

$\dot W_{in} + \dot m \cdot (h_{in}-h_{out}) = 0$

The properties of the fluid at entrance and exit are, respectively:

Inlet (Saturated Liquid)

$P = 20\,kPa$

$T = 60.06\,^{\textdegree}C$

$h = 251.42\,\frac{kJ}{kg}$

$s = 0.8320\,\frac{kJ}{kg\cdot K}$

Outlet (Subcooled Liquid)

$P = 6000\,kPa$

$T = 60.06\,^{\textdegree}C$

$h = 257.502\,\frac{kJ}{kg}$

$s = 0.8320\,\frac{kJ}{kg\cdot K}$

The power input to the pump is computed hereafter:

$\dot W_{in} = \left(45\,\frac{kg}{s} \right)\cdot \left(257.502\,\frac{kJ}{kg} -251.42\,\frac{kJ}{kg} \right)$

$\dot W_{in} = 273.69\,kW$

8 0

4 years ago

An uncovered swimming pool loses 1.0 inch of water off its 1,000 ft^2 surface each week due to evaporation. The heat of vaporiza

soldi70 [24.7K]

Answer:

The affirmation is true, the cover will be worth buying

Explanation:

The equation necessary to use is

E = m*cv,

Where

cv: the heat of vaporization.

Finding the rate at which the water evaporates (m^3/week).

The swimming pool loses water at 1 inch/week off its 1,000 ft^2

Than,

1000 ft² * 1 in/wk * 1 ft/12 in = 83.33 ft³/week

To obtains the rate of mass loss it is necessary to multiply it for the density of water

83.33 ft³/week * 62.4 lb/ft³ = 5200 lb/week

Knowing the vaporization heat it is possible to find the rate of heat which is leaving the swimming pool

5200 lb/week * 1050 BTU/lb = 5460000 btu/week

Over a 15-week period, the pool loses 81.9 million BTU.

Knowing the cost of energy to heat the pool is $10.00 per million btu

The price = $819

This way, the affirmation is true, the cover will be worth buying

3 0

3 years ago

If a torque of M = 300 N⋅m is applied to the flywheel, determine the force that must be developed in the hydraulic cylinder CD t

Licemer1 [7]

Answer:

Explanation:

The torque on the flywheel = 300 N-m

The force from the hydraulic cylinder will generate a moment on CA about point A.

The part of this moment that will be at point B about A must be proportional to the torque on the cylinder which is 300 N-m

we know that moment = F x d

where F is the force, and

d is the perpendicular distance from the turning point = 1 m

Equating, we have

300 = F x 1

F = 300 N this is the frictional force that stops the flywheel

From F = μN

where F is the frictional force

μ is the coefficient of static friction = 0.4

N is the normal force from the hydraulic cylinder

substituting, we have

300 = 0.4 x N

N = 300/0.4 = 750 N

This normal force calculated is perpendicular to CA. This actual force, is at 30° from the horizontal. To get the force from the hydraulic cylinder R, we use the relationship

N = R sin (90 - 30)

750 = R sin 60°

750 = 0.866R

R = 750/0.866 = <em>866.1 N</em>

3 0

3 years ago

Please calculate the current for the circuit below

Aleks04 [339]

Answer:

The answer is " $I = 0.0085106383 \ A$ "

Explanation:

Given:

$R= 470 \ \Omega \\\\V= 4 \ v$

Formula:

$\to V=IR\\\\\to I = \frac{V}{R}\\\\$

$= \frac{4}{470}\\\\ = 0.0085106383 \ A$

3 0

3 years ago