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Radda [10]
3 years ago
5

If the mass of an object increases, how is its acceleration affected, assuming the net force acting on the object remains the sa

me?
Physics
1 answer:
vovikov84 [41]3 years ago
7 0
Based on Newton's second law of motion, the net force applied to an object is equal to the product of the mass of the object and the acceleration it experiences. That is,
  
          F = ma

If we are to assume that the net force is constant and that the mass is increased, the acceleration should therefore decrease in order to make constant the value at the right-hand side of the equation. 
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Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500°C, and 80 m/s and the exit c
FrozenT [24]

Answer:

a) \Delta \dot K = 24.570\,kW, b) \dot W_{out} = 12729.15\,kW, c) A_{in} = 0.0136\,m^{2}

Explanation:

A turbine is a device which works usually in steady state and assumption of being adiabatic means no heat interactions between steam through turbine and surroudings and produce mechanical work from fluid energy. Changes in gravitational energy can be neglected. This system can be modelled after the First Law of Thermodynamics:

-\dot W_{out} + \dot m \cdot (h_{in} - h_{out}) + \frac{1}{2}\cdot \dot m \cdot (v_{in}^{2}-v_{out}^{2})   = 0

a) Change in kinetic energy

\Delta \dot K = \frac{1}{2}\cdot \dot m \cdot (v_{in}^{2} - v_{out}^{2})

\Delta \dot K = \frac{1}{2} \cdot \left(12.6\,\frac{kg}{s} \right) \cdot \left[\left(80\,\frac{m}{s} \right)^{2}-\left(50\,\frac{m}{s} \right)^{2}\right]

\Delta \dot K = 24570\,W

\Delta \dot K = 24.570\,kW

b) Power output

\dot W_{out} = \dot m \cdot (h_{in} - h_{out}) + \frac{1}{2}\cdot \dot m \cdot (v_{in}^{2}-v_{out}^{2})

\dot W_{out} = \left(12.6\,\frac{kg}{s}\right)\cdot \left(3446\,\frac{kJ}{kg} - 2437.7\,\frac{kJ}{kg} \right) + 24.570\,kW

\dot W_{out} = 12729.15\,kW

c) Turbine inlet area

Turbine inlet area can be found by using the following expressions:

\dot V_{in} = \dot m \cdot \nu_{in}

\dot V_{in} = \left(12.6\,\frac{kg}{s}\right) \cdot \left(0.086442\,\frac{m^{3}}{kg} \right)

\dot V_{in} = 1.089\,\frac{m^{3}}{s}

A_{in} = \frac{\dot V_{in}}{v_{in}}

A_{in} = \frac{1.089\,\frac{m^{3}}{s} }{80\,\frac{m}{s} }

A_{in} = 0.0136\,m^{2}

8 0
3 years ago
What best describes the difference between the second sound wave and the first?
Vikki [24]
Missing an attached sheet with the sound waves. Please add the attached sheet so I can properly help.

Thanks! :D
8 0
3 years ago
Air pressure is 1.0 · 105 N/m2, air density is 1.3 kg/m3, and the density of soft drinks is 1.0 · 103 kg/m3. If one blows carefu
natita [175]

Answer:

v = 27.456 m/s

Explanation:

The support pressure needed of the water in the straw can be calculated by the formula

Given that,

P = r*g*h

= 1000*9.8*0.05 Pa.= 490 Pa

This pressure is compensated by 0.5*r*v^2 of the air,

Hence,

0.5*1.3*v^2 = 490

velocity of air blown into the straw =

v = 27.456 m/s

8 0
3 years ago
What is the power of a parallel circuit with a resistance of 1,000 and a current of 0.03 A?
Kobotan [32]

i squared r = 0.03x0.03x1000=3x0.03x10=.9W

7 0
3 years ago
Ayuda por favor cual es el objetivo del ultimate?
postnew [5]

Answer: i d k

Explanation:

3 0
3 years ago
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