Was rachel and ross on a break

You’ve experienced convection cooling if you’ve ever extended your hand out the window of a moving vehicle or into a flowing wat

Anna35 [415]

Answer:

Condition A

Heat flux is 1400 W/M^2

Condition B

Heat flux is 12800 w/m^2

Explanation:

Given that:

$T_s$ is given as 30 degree celcius

condition A

Air temperature = - 5 degree c

convection coefficient h = 40 w/m^2. k

$heat\ flux = \frac{Q}{a}= h\Delta = 40{30 - (-5)} = 1400 w/m^2$

condition A

water temperature = 10 degree c

convection coefficient = 800 w/m^2.k

$heat\ flux = \frac{Q}{A} = H(\Delta} = 800\times (30-14) = 12800w/m^2$

7 0

2 years ago

List and describe three classifications of burns to the body.

DiKsa [7]

AnswerWhat Are the Classifications of Burns? Burns are classified as first-, second-, or third-degree, depending on how deep and severe they penetrate the skin's surface. First-degree burns affect only the epidermis, or outer layer of skin. The burn site is red, painful, dry, and with no blisters.

Explanation:

8 0

2 years ago

Read 2 more answers

A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16

abruzzese [7]

Answer:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The net work per cycle is 845.88 kJ/kg

The power developed in horsepower ≈ 45374 hP

Explanation:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The dimension of the cylinder bore diameter = 3.7 in. = 0.09398 m

Stroke length = 3.4 in. = 0.08636 m.

The volume of the cylinder v₁= 0.08636 ×(0.09398²)/4 = 5.99×10⁻⁴ m³

The clearance volume = 16% of cylinder volume = 0.16×5.99×10⁻⁴ m³

The clearance volume, v₂ = 9.59 × 10⁻⁵ m³

p₁ = 14.5 lbf/in.² = 99973.981 Pa

T₁ = 60 F = 288.706 K

$\dfrac{T_{2}}{T_{1}} = \left (\dfrac{v_{1}}{v_{2}} \right )^{K-1}$

Otto cycle T-S diagram

T₂ = 288.706* $6.25^{0.393}$ = 592.984 K

The maximum temperature = T₃ = 5200 R = 2888.89 K

$\dfrac{T_{3}}{T_{4}} = \left (\dfrac{v_{4}}{v_{3}} \right )^{K-1}$

T₄ = 2888.89 / $6.25^{0.393}$ = 1406.5 K

Work done, W = $c_v$ ×(T₃ - T₂) - $c_v$ ×(T₄ - T₁)

0.718×(2888.89 - 592.984) - 0.718×(1406.5 - 288.706) = 845.88 kJ/kg

The power developed in an Otto cycle = W×Cycle per second

= 845.88 × 2400 / 60 = 33,835.377 kW = 45373.99 ≈ 45374 hP.

8 0

3 years ago

Two production methods are being compared. One manual and the other automated. The manual method produces 10 pc per hour and req

Genrish500 [490]

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

6 0

3 years ago

A program is seeded with 30 faults. During testing, 21 faults are detected, 15 of which are seeded faults and 6 of which are ind

Vesna [10]

Answer:

Estimated number of indigenous faults remaining undetected is 6

Explanation:

The maximum likelihood estimate of indigenous faults is given by,

$N_F=n_F\times \frac{N_S}{n_S}$ here,

$n_F$ = the number of unseeded faults = 6

$N_S$ = number of seeded faults = 30

$n_s$ = number of seeded faults found = 15

So NF will be calculated as,

$N_F=6\times \frac{30}{15}=12$

And the estimate of faults remaining is $N_F-n_F$ = 12 - 6 = 6

8 0

2 years ago