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3 years ago

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A hot air balloon must be designed to support a basket, cords, and one person for a total payload weight of 1300 N plus the addi

tional weight of the balloon material itself. The balloon material has a mass of 60 g/m2. Ambient air is at 25 °C and 1 atm. The hot air inside the balloon is at 70 °C and 1 atm. Assuming that the balloon has a perfectly spherical shape, what balloon diameter will just support the total weight? Neglect the size of the hot air inlet vent and assume an ideal gas law (P = rhoRT) for the air.

1 answer:

RSB [31]3 years ago

3 0

Answer:

r = 4.44 m

Explanation:

For this exercise we use the Archimedes principle, which states that the buoyant force is equal to the weight of the dislodged fluid

B = ρ g V

Now let's use Newton's equilibrium relationship

B - W = 0

B = W

The weight of the system is the weight of the man and his accessories (W₁) plus the material weight of the ball (W)

σ = W / A

W = σ A

The area of a sphere is

A = 4π r²

W = W₁ + σ 4π r²

The volume of a sphere is

V = 4/3 π r³

Let's replace

ρ g 4/3 π r³ = W₁ + σ 4π r²

If we use the ideal gas equation

P V = n RT

P = ρ RT

ρ = P / RT

P / RT g 4/3 π r³ - σ 4 π r² = W₁

r² 4π (P/3RT r - σ) = W₁

Let's replace the values

r² 4π (1.01 10⁵ / (3 8.314 (70 + 273)) r - 0.060) = 13000

r² (11.81 r -0.060) = 13000 / 4pi

r² (11.81 r - 0.060) = 1034.51

As the independent term is very small we can despise it, to find the solution

r = 4.44 m

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Answer:

(1) Offer the right benefits

(2) Be transparent and open

Explanation:

There are two key strategies for retaining step are discussed below:

(1) Offer the right benefits: Benefits and perks can play a very good role in keeping employees happy, healthy and engaged. But benefits can go far beyond paid sick leaves and health coverage.

The company can give some financial awards to the employees which can help in exceeding performance and by this employees will stay in the company for a long period. Nearly 9 out of 10 companies will provide incentives and bonuses to the employees in coming next five years, according to the CTA.

(2) Be transparent and open: Creating open communication between the management and employees will help in the growth of the company as well as make the office atmosphere good. By this employees can share the ideas with the management and by these meetings, the employees can share their problems with management frankly will help employees to stay in the company for a longer period.

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3 years ago

A torsion-bar spring consists of a prismatic bar, usually of round cross section, that is twisted at one end and held fast at th

ra1l [238]

Answer:

d₁ = 0.29 in

d₂ = 0.505 in

Explanation:

Given:

T = 1500 lbf in

L = 10 in

x = 0.5 L = 5 in

$T_{1} =\frac{T(L-x)}{L} =\frac{1500*(10-5)}{10} =750lbfin$

First case: T = T₁ + T₂

T₂ = T - T₁ = 1500 - 750 = 750 lbf in

If the shafts are in series:

θ = θ₁ + θ₂

θ = ((T₁ * L₁)/GJ) + ((T₂ * L₂)/GJ)

Second case: If d₁ ≠ d₂

θ = ((T₁ * L₁)/GJ₁) + ((T₂ * L₂)/GJ₂) = 0 (eq. 1)

t₁ = t₂

$\frac{16T_{1} }{\pi d_{1}^{3} } =\frac{16T_{2} }{\pi d_{2}^{3} }$ (eq. 2)

T₁ + T₂ = 1500 (eq. 3)

θ₁ first case = θ₁ second case

Replacing:

$\frac{750*5}{G(\frac{\pi }{32})*0.5^{4} } =\frac{T_{1}*3.7 }{G(\frac{\pi }{32})*d_{1} ^{4} }\\T_{1} =16216d_{1} ^{4}$

The same way to θ₂:

$\frac{750*5}{G(\frac{\pi }{32})*0.5^{4} } =\frac{T_{2}*6.3 }{G(\frac{\pi }{32})*d_{2} ^{4} } \\T_{2} =9523.8d_{2} ^{4}$

From equation 2, we have:

d₁ = 0.587 * d₂

From equation 3, we have:

d₂ = 0.505 in

d₁ = 0.29 in

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A blue-green photon (λ = 488 nm ) is absorbed by a free hydrogen atom, initially at rest. What is the recoil speed of the hydrog

Natalka [10]

Answer:

The recoil speed is $2.207\times 10^{4} m/s$

Solution:

Wavelength of a blue-green photon, $\lambda_{BG} = 488 nm = 488\times 10^{- 9} m$

Now, the energy associated with the blue-green photon:

$E_{BG} = \frac{hc}{\lambda_{BG}}$

where

h = Planck's constant

C = speed of light ion vacuum

$E_{BG} = \frac{6.626\times 10^{- 34}\times 3\times 10^{8}}{488\times 10^{- 9}}$

$E_{BG} = 4.07\times 10^{- 19} J$

Also, we know that the recoil speed can be calculated by the KInetic energy which is equal to the Energy of the blue-green photon:

$KE_{H} =\frac{1}{2}m_{p}v_{H}$

where

$v_{H}$ = velocity of Hydrogen atom

$m_{p} = 1.67\times 10^{- 27} kg$ = mass of H-atom

Now,

$KE_{H} =\frac{1}{2}m_{p}(v_{H})^{2}$

$4.07\times 10^{- 19} =\frac{1}{2}\times 1.67\times 10^{- 27}\times (v_{H})^{2}$

$v_{H} = \sqrt(4.87\times 10^{8}) = 2.207\times 10^{4} m/s$

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For the question above, here is the equation to follow:
<span>F = mgsinα = Wsinα
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<span>Sin30 * 200N = 100 N
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Answer:

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