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Doss [256]
3 years ago
8

What is the potential consequence of online impulsive buying?

Engineering
2 answers:
cricket20 [7]3 years ago
7 0
Probably B because I impulsively shop online and b is true for me. I’m not 100% sure though.
Nastasia [14]3 years ago
3 0
I’m pretty sure it’s B
You might be interested in
An actual vapour compression system comprises following process represents a. 1-2 Compression process b. 2-3 Condens 1 (or heat
Gemiola [76]

Answer:

Explanation:

The deatailed diagram of VCRS is given below such

1-2=Isentropic compression in which temperature increases at constant entropy

2-3=Isobaric heat rejection i.e. heat rejected at constant pressure(condensation)

3-4=Irreversible expansion or throttling in which enthalpy remains constant

4-1=Isobaric heat addition(Evaporation)

4 0
3 years ago
One of the key characteristics of ________ sessions is that no idea should be immediately accepted or rejected. prototype alpha
Alex_Xolod [135]

Answer:

brainstorming

Explanation:

8 0
2 years ago
Argon is compressed in a polytropic process with n = 1.2 from 100 kPa and 30°C to 1200 kPa in a piston–cylinder device. Determin
gulaghasi [49]

Answer:

<em>181 °C</em>

<em></em>

Explanation:

Initial pressure P_{1} = 100 kPa

Initial temperature T_{1} = 30 °C = 30 + 273 K = 303 K

Final pressure P_{2} = 1200 kPa

Final temperature T_{2} = ?

n = 1.2

For a polytropic process, we use the relationship

(T_{2}/T_{1} ) = (P_{2}/P_{1})^γ

where γ = (n-1)/n

γ = (1.2-1)/1.2 = 0.1667

substituting into the equation, we have

(T_{2}/303) = (1200/100)^0.1667

T_{2}/303 = 12^0.1667

T_{2}/303 = 1.513

T_{2} = 300 x 1.513 = 453.9 K

==> 453.9 - 273 = 180.9 ≅ <em>181 °C</em>

5 0
2 years ago
A car generator turns at 400 rpm (revolutions per minute) when the engine is idling. It has a rectangular coil with 300 turns of
larisa86 [58]

Answer:

The field strength needed is 0.625 T

Explanation:

Given;

angular frequency, ω = 400 rpm = (2π /60) x (400) = 41.893 rad/s

area of the rectangular coil, A =  L x B = 0.0611 x 0.05 = 0.003055 m²

number of tuns of the coil, N = 300 turns

peak emf = 24 V

The peak emf is given by;

emf₀ = NABω

B = (emf₀ ) / (NA ω)

B = (24) / (300 x 0.003055 x 41.893)

B = 0.625 T

Therefore, the field strength needed is 0.625 T

4 0
3 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
2 years ago
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