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ziro4ka [17]
2 years ago
11

In a cellular phone system, a mobile phone must be paged to receive a phone call. However, paging attempts don’t always succeed

because the mobile phone may not receive the paging signal clearly. Consequently, the system will page a phone up to three times before giving up. If the results of all paging attempts are independent and a single paging attempt succeeds with probability 0.8, sketch a probability tree for this experiment and find the probability P[F] that the phone receives the paging signal clearly.
Engineering
1 answer:
Alina [70]2 years ago
7 0

Answer:

The correct response will be "0.992". The further explanation to the following question is given below.

Explanation:

The probability that paging would be beneficial becomes 0.8  

Effective paging at the very first attempted is 0.8

On the second attempt the success probability will be:

⇒  0.2\times 0.8

⇒  0.16

On the third attempt the success probability will be:

⇒  0.2\times 0.2\times 0.8

⇒  0.032

So that the success probability will be:

⇒  0.8 + 0.16 + 0.032

⇒  0.992

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A closed system consisting of 4 lb of a gas undergoes a process during which the relation between pressure and volume is pVn 5 c
gayaneshka [121]

Answer:

V1=5<u>ft3</u>

<u>V2=2ft3</u>

n=1.377

Explanation:

PART A:

the volume of each state is obtained by multiplying the mass by the specific volume in each state

V=volume

v=especific volume

m=mass

V=mv

state 1

V1=m.v1

V1=4lb*1.25ft3/lb=5<u>ft3</u>

state 2

V2=m.v2

V2=4lb*0.5ft3/lb=  <u> 2ft3</u>

PART B:

since the PV ^ n is constant we can equal the equations of state 1 and state 2

P1V1^n=P2V2^n

P1/P2=(V2/V1)^n

ln(P1/P2)=n . ln (V2/V1)

n=ln(P1/P2)/ ln (V2/V1)

n=ln(15/53)/ ln (2/5)

n=1.377

3 0
2 years ago
2. A ¼ in. diameter rod must be machined on a lathe to a smaller diameter for use as a specimen in a tension test. The rod mater
gladu [14]

Answer:

we use

sigma=force /area\\A=\pi r^2\\D=2*r

5 0
2 years ago
A pair of spur gears with 20 degree pressure angle, full-depth, involute teeth transmits 65 hp. The pinion is mounted on a shaft
vfiekz [6]

Answer:

See explaination

Explanation:

a) The pitch circle diameter of pinion in inches is given by

Dp=Np/P

Where

Np= No. of teeth in pinion = 26

P =diametral pitch= 6

Hence

Dp= 26/6 = 4.333 in

Pinion angular speed\omega _{p} =1250 rpm = 130.9 rad/s

Therefore pitch line speed

V=Dp/2\omega _{p} = 4.333/2x130.9

= 283.62 in/s

V= 23.63 ft/s

V=1418 ft/min

b) The pitch circle diameter og gear is given by

Dg= Ng/P= 48/6 = 8 in

The center distance is given by

C=(Dp+Dg)/2

= (4.333+8)/2

C= 6.167 in

c) The torque on the pinion is given by

Tp= P/\omega _{p}

Where

P = transmitted power, =65 hp = 65x550= 35750 lt-lb/s

Tp= 35750/130.9

= 273.1 ft-lb

d) Speed ratio is given as

R=Ng/Np= 48/26 = 1.8461

Hence speed of gear is

\omega _{g}=\frac{\omega _{p}}{R}

= 130.9/1.8461

= 70.9 rad/s

Therefore torque on gear is given as

Tg= P/\omega _{g} = 35750/70.9

= 504.2 ft-lb

e) Assuming transmission eficiency of 100%

Output hp=input hp= 65 hp

f) Tangential force on gear teeth is given by

Fgt= Tg/(Dg/2)

= 504.2x2/8

= 126.05 lb

g) Radial force on ger teeth is given as

Fgr= Fgt tan\phi

Where

\phi is pressure angle = 200

Hence

Fgr= 126.05tan200

= 45.88 lb

h) The normal force on gear teeth is given as

F=Fgt/cos\phi

= 126.05/cos200

= 134.14 lb

4 0
2 years ago
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Nikitich [7]

Answer:

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Explanation:

70 \ kPa  \ \ and \ \ 100^{\circ}C \\\\s_i= 7.56162\ \frac{kJ}{kgK}\\\\u_i= 2509.39 \ \frac{kJ}{kg}\\\\

The method is isentropic since the cylinders are shielded.

Calculating the work:

w= u_2-u_i \\\\

   = 3418.7728-2509.38 \\\\=909.3928 \ KJ

3 0
2 years ago
Consider a normal shock wave in air. The upstream conditions are given by M1=3, p1 = 1 atm, and r1 = 1.23 kg/m3. Calculate the d
mart [117]

Answer and Explanation:

The answer is attached below

7 0
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