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Mrrafil [7]
3 years ago
9

What’s cat plus cat ?

Engineering
2 answers:
masya89 [10]3 years ago
8 0
Cat + Cat = kitten/baby cat
iris [78.8K]3 years ago
5 0
Uhmmmmm, a kitten...? Lol
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Thermal energy generated by the electrical resistance of a 5-mm-diameter and 4-m-long bare cable is dissipated to the surroundin
ella [17]

Answer:

surface temperature = 128.74⁰c

Explanation:

Given data

diameter of cable = 5 mm = 0.005 m

length of cable = 4 m

T∞ ( surrounding temperature ) = 20⁰c

voltage drop across cable ( dv )= 60 V

current across cable = 1.5 A

attached to this answer is the comprehensive analysis and solution to the problem.

The assumption made is not a good one since the calculated Ts ( surface temperature ) is very much different from the assumed Ts

6 0
2 years ago
In a simple ideal Rankine cycle, water is used as the working fluid. The cycle operates with pressures of 2000 psi in the boiler
weqwewe [10]

Answer:

Explanation:

The pressures given are relative

p1 = 2000 psi

P1 = 2014 psi = 13.9 MPa

p2 = 4 psi

P2 = 18.6 psi = 128 kPa

Values are taken from the steam pressure-enthalpy diagram

h2 = 2500 kJ/kg

If the output of the turbine has a quality of 85%:

t2 = 106 C

I consider the expansion in the turbine to adiabatic and reversible,  therefore, isentropic

s1 = s2 = 6.4 kJ/(kg K)

h1 = 3500 kJ/kg

t2 = 550 C

The work in the turbine is of

w = h1 - h2 = 3500 - 2500 = 1000 kJ/kg

The thermal efficiency of the cycle depends on the input heat.

η = w/q1

q1 is  not a given, so it cannot be calculated.

3 0
3 years ago
Carbon dioxide steadily flows into a constant pressure heater at 300 K and 100 kPa witha mass flow rate of 9.2 kg/s. Heat transf
docker41 [41]

Answer:

Carbon dioxide temperature at exit is 317.69 K

Carbon dioxide flow rate at heater exit is 20.25 m³/s

Explanation:

Detailed steps are attached below.

8 0
2 years ago
A fatigue test is performed on 69 rotating specimens made of 5160H steel. The measured number of cycles to failure (L in kcycles
tensa zangetsu [6.8K]

Answer:

(a) Mean = 122.9, σ = 30.071

(b) No. of failed specimens at less than 115k cycles are 27.

(c) μ = 39.07

Explanation:

We are given:

L  60  70  80  90  100  110  120  130  140  150  160  170  180  190  200  210

f    2     1    3     5     8     12     6     10     8     5     2      3      2      1       0      1

(a) First we need to calculate the mean and standard deviation. The formula for calculating mean is:

Mean = ∑fx/∑f

And for standard deviation we have:

S.D. = √Var

Var = ∑fx²/∑f - (Mean)²

∑fx = (2*60) + (1*70) + (3*80) + (5*90) + (8*100) + (12*110) + (6*120) + (10*130) + (8*140) + (5*150) + (2*160) + (3*170) + (2*180) + (1*190) + (0*200) + (1*210)

         = 120 + 70 + 240 + 450 + 800 + 1320 + 720 + 1300 + 1120 + 750 + 320 + 510 + 360 + 190 + 0 + 210

∑fx = 8480

Mean = ∑fx/∑f

          = 8480/69

Mean = 122.9  

∑fx² = (2*60²) + (1*70²) + (3*80²) + (5*90²) + (8*100²) + (12*110²) + (6*120²) + (10*130²) + (8*140²) + (5*150²) + (2*160²) + (3*170²) + (2*180²) + (1*190²) + (0*200²) + (1*210²)

   =7200+4900+19200+40500+80000+145200+86400+169000+156800+112500+51200+86700+64800+36100+0+44100

∑fx² = 1104600

Var = ∑fx²/∑f - (Mean)²

     = 1104600/69 - (122.9)²

     = 16008.69565 - 15104.41

Var = 904.2856

S.D = √Var

σ = √904.2856

σ = 30.071

(b) Let X be the number of failed specimen.

We will use the z-score to calculate the probability. The formula for z-score is:

z = (X-μ)/σ

P(X<115) = P(z<(115-122.9)/30.071)

              = P(z<-0.26)

Using the normal distribution probability table, we can compute the value of  P(z<-0.26).

P(X<115) = 0.3974

So, no. of failed specimens at less than 115k cycles are: 0.3974*69 = 27 specimens

(c) σ = 30.071

P(x<115) = 0.99

P(z<(115-μ)/30.071) = 0.99

From the normal distribution table we find that 0.99 lies between the z values 2.52 and 2.33. Hence, we get 2.525 as the z-value at which the probability is 0.99.

z = (x-μ)/σ

2.525 = (115 - μ)/30.071

75.93 = 115 - μ

μ = 115 - 75.93

μ = 39.07

4 0
3 years ago
Disconnecting means shall be capable of being locked in the open position. The provisions for locking do not have to reamin in p
Lorico [155]
The answer is a.true
7 0
3 years ago
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