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Bumek [7]
3 years ago
6

A pointer is spun on a fair wheel of chance having its periphery labeled Trom 0 to 100. (a) Whhat is the sample space for this e

xperiment? (b)What is the probability that the pointer will stop between 20 and 35? (c) What is the probability that the wheel will stop on 58?
​
Engineering
2 answers:
liubo4ka [24]3 years ago
7 0

Answer:

A pointer is spun on a fair wheel of chance having its periphery labeled Trom 0 to 100. (a) Whhat is the sample space for this experiment? (b)What is the probability that the pointer will stop between 20 and 35? (c) What is the probability that the wheel will stop on 58?

​

Explanation:

thats all you said

Wewaii [24]3 years ago
5 0

Answer:

hii my name is RAGHAV what is your name

Explanation:

this question is which chapter

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1. (5 pts) An adiabatic steam turbine operating reversibly in a powerplant receives 5 kg/s steam at 3000 kPa, 500 °C. Twenty per
KiRa [710]

Answer:

temperature of first extraction 330.8°C

temperature of second extraction 140.8°C

power output=3168Kw

Explanation:

Hello!

To solve this problem we must use the following steps.

1. We will call 1 the water vapor inlet, 2 the first extraction at 100kPa and 3 the second extraction at 200kPa

2. We use the continuity equation that states that the mass flow that enters must equal the two mass flows that leave

m1=m2+m3

As the problem says, 20% of the flow represents the first extraction for which 5 * 20% = 1kg / s

solving

5=1+m3

m3=4kg/s

3.

we find the enthalpies and temeperatures in each of the states, using thermodynamic tables

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties

4.we find the enthalpy and entropy of state 1 using pressure and temperature

h1=Enthalpy(Water;T=T1;P=P1)

h1=3457KJ/kg

s1=Entropy(Water;T=T1;P=P1)

s1=7.234KJ/kg

4.

remembering that it is a reversible process we find the enthalpy and the temperature in the first extraction with the pressure 1000 kPa and the entropy of state 1

h2=Enthalpy(Water;s=s1;P=P2)

h2=3116KJ/kg

T2=Temperature(Water;P=P2;s=s1)

T2=330.8°C

5.we find the enthalpy and the temperature in the second extraction with the pressure 200 kPav y the entropy of state 1

h3=Enthalpy(Water;s=s1;P=P3)

h3=2750KJ/kg

T3=Temperature(Water;P=P3;s=s1)

T3=140.8°C

6.

Finally, to find the power of the turbine, we must use the first law of thermodynamics that states that the energy that enters is the same that must come out.

For this case, the turbine uses a mass flow of 5kg / s until the first extraction, and then uses a mass flow of 4kg / s for the second extraction, taking into account the above we infer the following equation

W=m1(h1-h2)+m3(h2-h3)

W=5(3457-3116)+4(3116-2750)=3168Kw

7 0
3 years ago
A Carnot heat engine operates between 1000 deg F and 50 deg F, producing 120 BTU of work. What is the heat input to the engine?
Mazyrski [523]

Answer:

184.6 BTU

Explanation:

The thermal efficiency for a Carnot cycle follows this equation:

η = 1 - T2/T1

Where

η: thermal efficiency

T1: temperature of the heat source

T2: temperature of the heat sink

These temperatures must be in absolute scale:

1000 F = 1460 R

50 F = 510 R

Then

η = 1 - 510/1460 = 0.65

We also know that for any heat engine:

η = L / Q1

Where

L: useful work

Q1: heat taken from the source

Rearranging:

Q1 = L / η

Q1 = 120 / 0.65 = 184.6 BTU

4 0
4 years ago
Calculate the volume of a hydraulic accumulator capable of delivering 5 liters of oil between 180 and 80 bar, using as a preload
Vinil7 [7]

Answer:

1) V_o = 10 liters

2) V_o = 12.26 liters

Explanation:

For isothermal process n =1

V_o =\frac{\Delta V}{(\frac{p_o}{p_1})^{1/n} -(\frac{p_o}{p_2})^{1/n}}

V_o  = \frac{5}{[\frac{72}{80}]^{1/1} -[\frac{72}{180}]^{1/1}}

V_o = 10 liters

calculate pressure ratio to determine correction factor

\frac{p_2}{p_1} =\frac{180}{80} = 2.25

correction factor for calculate dpressure ration  for isothermal process is

c1 = 1.03

actual \ volume = c1\times 10 = 10.3 liters

b) for adiabatic process

n =1.4

volume of hydraulic accumulator is given as

V_o =\frac{\Delta V}{[\frac{p_o}{p_1}]^{1/n} -[\frac{p_o}{p_2}]^{1/n}}

V_o  = \frac{5}{[\frac{72}{80}]^{1/1.4} -[\frac{72}{180}]^{1/1.4}}

V_o = 12.26 liters

calculate pressure ratio to determine correction factor

\frac{p_2}{p_1} =\frac{180}{80} = 2.25

correction factor for calculate dpressure ration  for isothermal process is

c1 = 1.15

actual \volume = c1\times 10 = 11.5 liters

8 0
3 years ago
Activity
Sphinxa [80]

Answer:

yo do you still need an

answer

Explanation:

6 0
4 years ago
What is the least common denominator for the fractions 16and34?
mihalych1998 [28]

Answer:

272

Explanation:

3 0
3 years ago
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