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konstantin123 [22]
3 years ago
13

When a force causes an object to move a distance ______ is done

Physics
1 answer:
Mumz [18]3 years ago
4 0
Work. But only if the vector of the force is in the same direction as when it moved.
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Cool facts about satellites ???
svetoff [14.1K]
Satellites travel at 18,000 miles per hour. ...

A satellite gets better fuel economy than a Prius.

If you put all of the data that our satellites collect in a year on DVDs, it would form a stack nearly 4 times the height of the Empire State Building.

There are over 2,500 satellites in orbit around the Earth.
7 0
3 years ago
What is potential energy?
iVinArrow [24]
energy associate with position or shape
5 0
4 years ago
Steam enters an adiabatic turbine steadily at 7 MPa, 5008C, and 45 m/s, and leaves at 100 kPa and 75 m/s. If the power output of
marusya05 [52]

Answer:

a) \dot m = 6.878\,\frac{kg}{s}, b) T = 104.3^{\textdegree}C, c) \dot S_{gen} = 11.8\,\frac{kW}{K}

Explanation:

a) The turbine is modelled by means of the First Principle of Thermodynamics. Changes in kinetic and potential energy are negligible.

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

The mass flow rate is:

\dot m = \frac{\dot W_{out}}{h_{in}-h_{out}}

According to property water tables, specific enthalpies and entropies are:

State 1 - Superheated steam

P = 7000\,kPa

T = 500^{\textdegree}C

h = 3411.4\,\frac{kJ}{kg}

s = 6.8000\,\frac{kJ}{kg\cdot K}

State 2s - Liquid-Vapor Mixture

P = 100\,kPa

h = 2467.32\,\frac{kJ}{kg}

s = 6.8000\,\frac{kJ}{kg\cdot K}

x = 0.908

The isentropic efficiency is given by the following expression:

\eta_{s} = \frac{h_{1}-h_{2}}{h_{1}-h_{2s}}

The real specific enthalpy at outlet is:

h_{2} = h_{1} - \eta_{s}\cdot (h_{1}-h_{2s})

h_{2} = 3411.4\,\frac{kJ}{kg} - 0.77\cdot (3411.4\,\frac{kJ}{kg} - 2467.32\,\frac{kJ}{kg} )

h_{2} = 2684.46\,\frac{kJ}{kg}

State 2 - Superheated Vapor

P = 100\,kPa

T = 104.3^{\textdegree}C

h = 2684.46\,\frac{kJ}{kg}

s = 7.3829\,\frac{kJ}{kg\cdot K}

The mass flow rate is:

\dot m = \frac{5000\,kW}{3411.4\,\frac{kJ}{kg} -2684.46\,\frac{kJ}{kg}}

\dot m = 6.878\,\frac{kg}{s}

b) The temperature at the turbine exit is:

T = 104.3^{\textdegree}C

c) The rate of entropy generation is determined by means of the Second Law of Thermodynamics:

\dot m \cdot (s_{in}-s_{out}) + \dot S_{gen} = 0

\dot S_{gen}=\dot m \cdot (s_{out}-s_{in})

\dot S_{gen} = (6.878\,\frac{kg}{s})\cdot (7.3829\,\frac{kJ}{kg\cdot K} - 6.8000\,\frac{kJ}{kg\cdot K} )

\dot S_{gen} = 11.8\,\frac{kW}{K}

4 0
4 years ago
Jin walked 4 km on a straight path to get to the sandwich shop. He traveled 30° south of east. What is the southern component of
mylen [45]
Here, we can use the resolution of vectors to find the component in the y-axis of the distance traveled by Jin. The southern component is given by:
s = dsin(∅), where s is the southern component, d is the distance traveled and ∅ is the angle below east at which it was traveled.
s = 4sin(30)
s = 2 km
8 0
3 years ago
Read 2 more answers
A 2550 pound roller coaster starts from rest and is launched such that it crests a 119 ft high hill with a speed of 57 mph. The
Feliz [49]

To solve this problem we will apply the concepts related to energy conservation. For this purpose we will have that all the changes occurred in the energy change will be equivalent to the change in the potential and kinematic energies of the body. At the same time we will consider that the change in the final energy of the system will be reflected in the work of the system, therefore,

\Delta E = KE+PE

E_i - Fd = \frac{1}{2}mv^2+Wh

Here,

F = Force

m = mass

v = Velocity

h = Height

d = Distance

W = mg \rightarrow \text{Weight/Force Weight}

E_i =\frac{1}{2}mv^2+Wh+Fd

E_i = \frac{1}{2} (\frac{W}{g})v^2 +Wh+fd

Replacing we have,

E_i = \frac{1}{2} (\frac{2550pounds}{32.2ft/s^2})(57mi/h(\frac{1.467ft/s}{1mi/h}))+(2550pounds)(119ft)+(100pounds)(583ft)

E_i =365061ft\cdot lb

Therefore the launch energy is 365061ft-lb

3 0
4 years ago
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