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4 years ago

14

If a master cylinder piston applies pressure to a brak piston that has twice as much square area what will be the force from the

second piston

1 answer:

sp2606 [1]4 years ago

5 0

Explanation:

Pressure = force/area

Pressure stays the same.

If the area is doubled the force is doubled.

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Minchanka [31]

Answer:

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3 years ago

What should be done to denote the end of a topic?

damaskus [11]

Answer:

Include a topic sentence. Conclusions should always begin with a topic sentence.

Use your introductory paragraph as a guide.

Summarize the main ideas.

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Explanation:

Hope this help

8 0

3 years ago

Heat is applied to a rigid tank containing water initially at 200C, with a quality of 0.25, until the pressure reaches 8 MPa. De

creativ13 [48]

Answer:1747.53KJ/kg

Explanation:

Given data

Initial Temprature $\left ( T_i\right )=200^{\circ}$

Quality $\left ( \chi\right )=0.25$

Final pressure $\left ( P_2\right )=8MPa$

Given tank is rigid therefore specific Volume remains same

now calculating properties at $200^{\circ}$

$\nu _f=0.001156m^3/kg$

$\nu {fg}=0.126434m^3/kg$

$h_f=852.26KJ/kg$

$h_{fg}=1939.8KJ/kg$

Now Specific volume of steam at $200^{\circ}$

$\nu _1=\nu _f+\chi \times \nu {fg}$

$\nu _1=0.001156+0.25\times 0.126434$

$\nu _1=0.03276m^3/kg$

$h_1=h_f+\chi \times h_{fg}$

$h_1=852.26+0.25\times 1939.8$

$h_1=1337.21 KJ/kg$

Now $\nu_1=\nu_2$

$0.03276=\nu_2$

At 8 MPa and $\nu =0.03276m^3/kg$

$\nu _g=0.02356$

Since $\nu _2\ is\ greater\ than \nu _g$

therefore final state of steam is superheated at $T=381^{\circ}$

at this state $h_2=3084.74 KJ/kg$ (obtained from steam table)

Therefore heat added

$Q=h_2-h_1=3084.74-1337.21=1747.53 KJ/kg$

3 0

3 years ago

What is another term for an actuator of a control system?

aliya0001 [1]

I believe it is D, System Controller.
It is responsible for opening things in the body like a valve. I also did my research and had a little help on this.

3 0

3 years ago

Un gas está confinado por un pistón en un cilindro con un diámetro de 0.57 m; sobre el pistón se pone un peso. La masa del pistó

saw5 [17]

Answer:

a) La fuerza ejercida sobre el gas por la atmósfera, el pistón y el peso es 27543.428 newtons.

b) La presión del gas es 107.939 kilopascales.

Explanation:

b) Después de leer el enunciado, comprendemos que la presión total ( $p_{T}$ ), medido en kilopascales, sobre el gas confinado es la suma de las presiones ejercidas por la atmósfera ( $p_{atm}$ ), el pistón ( $p_{P}$ ) y el peso ( $p_{W}$ ), medidos en kilopascales. Es decir:

$p_{T} = p_{atm}+p_{P}+p_{W}$ (1)

Por la definición de presión, ampliamos la ecuación como sigue:

$p_{T} =p_{atm} + \frac{4\cdot (m_{P}+m_{W})\cdot g}{1000\cdot \pi\cdot D^{2}}$ (2)

Donde:

$m_{P}$ - Masa del pistón, medido en kilogramos.

$m_{W}$ - Masa del peso, medido en kilogramos.

$g$ - Aceleración gravitacional, medida en metros por segundo al cuadrado.

$D$ - Diámetro del pistón, medido en metros.

Si sabemos que $m_{P}+m_{W} = 160\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $D = 0.57\,m$ y $p_{atm} = 101.79\,kPa$ , entonces la presión del gas es:

$p_{T} = 101.79\,kPa+\frac{4\cdot (160\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{1000\cdot \pi \cdot (0.57\,m)^{2}}$

$p_{T} \approx 107.939\,kPa$

La presión del gas es 107.939 kilopascales.

a) La fuerza en newtons ejercida sobre el gas se obtiene al multiplicar el resultado anterior por el área del cilindro:

$F = 1000\cdot p_{T}\cdot \left(\frac{\pi}{4}\cdot D^{2} \right)$ (2)

Donde $F$ es la fuerza, medida en newtons.

Si sabemos que $p_{T} \approx 107.939\,kPa$ y $D = 0.57\,m$ , entonces la fuerza es:

$F = 1000\cdot (107.939\,kPa)\cdot \left(\frac{\pi}{4} \right)\cdot (0.57\,m)^{2}$

$F = 27543.428\,N$

La fuerza ejercida sobre el gas por la atmósfera, el pistón y el peso es 27543.428 newtons.

5 0

3 years ago