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3 years ago

The Stefan-Boltzmann law can be employed to estimate the rate of radiation of energy H from a surface, as in

Engineering

1 answer:

Mazyrski [523]3 years ago

5 0

Explanation:

H = Aeσ^4

Using the stefan Boltzmann law

When we differentiate

dH/dT = 4AeσT³

dH/dT = 4(0.15)(0.9)(5.67)(10^-8)(650)³

= 8.4085

Exact error = 8.4085x20

= 168.17

H(650) = 0.15(0.9)(5.67)(10^-8)(650)⁴

= 1366.376watts

Verifying values

H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(670)⁴

= 1542.468

H(T+ΔT) = 0.15(0.9)(5.67)(10^-8)(630)⁴

= 1205.8104

Error = 1542.468-1205.8104/2

= 168.329

ΔT = 40

H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(690)⁴

= 1735.05

H(T-ΔT) = 0.15(0.9)(5.67)(10^-8)(610)⁴

= 1735.05-1059.83/2

= 675.22/2

= 337.61

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Why did Percy most likely send Medusa's head to Olympus when he knew this would upset the gods?

Maksim231197 [3]

Answer: To show his father what he had done

Explanation:

Percy Jackson is the son of the great Greek god of the seas, Poseidon. He did not however, know this for a long time. After he finds out, his life becomes more exciting as he embarks on many adventures.

According to Grover who is his best friend and guardian, Percy sent Medusa's head to Mount Olympus to show his father what he had done so that Poseidon would acknowledge and be proud of him.

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2 years ago

What's the relationship between energy and time<br>

boyakko [2]

Answer:

The relationship between power, energy, and time can be described by the following equation : P = Δ E s y s Δ t. P is the average power output, measured in watts (W) ΔEsys is the net change in energy of the system in joules (J) - also known as work. Δt is the duration - how long the energy use takes - measured in seconds (s).

Explanation:

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2 years ago

What is chemical engineering

viktelen [127]

Answer:

Chemical engineering is the branch of engineering that deals with chemical production and the manufacture of products through chemical processes

Explanation:

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3 years ago

The compressed-air tank has an inner radius r and uniform wall thickness t. The gage pressure inside the tank is p and the centr

Sedaia [141]

Answer:

Explanation:

Given that:

The Inside pressure (p) = 1402 kPa

= 1.402 × 10³ Pa

Force (F) = 13 kN

= 13 × 10³ N

Thickness (t) = 18 mm

= 18 × 10⁻³ m

Radius (r) = 306 mm

= 306 × 10⁻³ m

Suppose we choose the tensile stress to be (+ve) and the compressive stress to be (-ve)

Then;

the state of the plane stress can be expressed as follows:

$(\sigma_ x) = \dfrac{Pd}{4t}+ \dfrac{F}{2 \pi rt}$

Since d = 2r

Then:

$(\sigma_ x) = \dfrac{Pr}{2t}+ \dfrac{F}{2 \pi rt}$

$(\sigma_ x) = \dfrac{1402 \times 306 \times 10^3}{2(18)}+ \dfrac{13 \times 10^3}{2 \pi \times 306\times 18 \times 10^{-3} \times 10^{-3}}$

$(\sigma_ x) = \dfrac{429012000}{36}+ \dfrac{13000}{34607.78467}$

$(\sigma_ x) = 11917000.38$

$(\sigma_ x) = 11.917 \times 10^6 \ Pa$

$(\sigma_ x) = 11.917 \ MPa$

$\sigma_y = \dfrac{pd}{2t} \\ \\ \sigma_y = \dfrac{pr}{t} \\ \\ \sigma _y = \dfrac{1402\times 10^3 \times 306}{18} \ N/m^2 \\ \\ \sigma _y = 23.834 \times 10^6 \ Pa \\ \\ \sigma_y = 23.834 \ MPa$

When we take a look at the surface of the circular cylinder parabolic variation, the shear stress is zero.

Thus;

$\tau _{xy} =0$

3 0

3 years ago

9. A box contains (4) red balls, and (7) white balls ,we draw( two) balls with return , find 1. Show the sample space & n(s)

zzz [600]

Answer:

The answers to your questions are given below.

Explanation:

The following data were obtained from the question:

Red (R) = 4

White (W) = 7

1. Determination of the sample space, S.

The box contains 4 red balls and 7 white balls. Therefore, the sample space (S) can be written as follow:

S = {R, R, R, R, W, W, W, W, W, W, W}

nS = 11

2. Determination of the probability of all results that appeared in the sample space.

From the question, we were told that the two balls was drawn with return. There, the probability of all results that appeared in the sample space can be given as follow:

i. Probability that the first draw is red and the second is also red.

P(R1) = nR/nS

Red (R) = 4

Space space (S) = 11

P(R1) = nR/nS

P(R1) = 4/11

P(R2) = nR/nS

P(R2) = 4/11

P(R1R2) = P(R1) x P(R2)

P(R1R2) = 4/11 x 4/11

P(R1R2) = 16/121

Therefore, the Probability that the first draw is red and the second is also red is 16/121.

ii. Probability that the first draw is red and the second is white.

Red (R) = 4

White (W) = 7

Space space (S) = 11

P(R) = nR/nS

P(R) = 4/11

P(W) = nW/nS

P(W) = 7/11

P(RW) = P(R) x P(W)

P(RW) = 4/11 x 7/11

P(RW) = 28/121

Therefore, the probability that the first draw is red and the second is white is 28/121.

iii. Probability that the first draw is white and the second is also white.

White (W) = 7

Space space (S) = 11

P(W1) = nW/nS

P(W1) = 7/11

P(W2) = nW/n/S

P(W2) = 7/11

P(W1W2) = P(W1) x P(W2)

P(W1W2) = 7/11 x 7/11

P(W1W2) = 49/121

Therefore, the probability that the first draw is white and the second is also white is 49/121.

iv. Probability that the first draw is white and the second is red.

Red (R) = 4

White (W) = 7

Space space (S) = 11

P(W) = nW/nS

P(W) = 7/11

P(R) = nR/nS

P(R) = 4/11

P(WR) = P(W) x P(R)

P(WR) = 7/11 x 4/11

P(WR) = 28/121

Therefore, the probability that the first draw is white and the second is red is 28/121.

7 0

3 years ago