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Vsevolod [243]
3 years ago
6

An airliner is flying at 34,000 ft cruise altitude on a standard day. Calculate the pressure difference between the cabin and th

e atmosphere if the cabin is pressurized to an altitude of 8,000 ft. Assume standard day temperature in the cabin
Engineering
1 answer:
nadya68 [22]3 years ago
7 0

Answer:

\Delta P=61,952.8\ lb/ft^2

Explanation:

Given

Airline flying at 34,000 ft.

Cabin pressurized to an altitude 8,000 ft.

We know that at standard condition ,density of air

\rho =0.074\ lb/ft^3

We know that pressure difference    

ΔP=ρ g ΔZ

Here ΔZ=34,000-8,000  ft

        ΔZ=26,000 ft

g= 32.2\ ft/s^2

ΔP=0.074 x 32.2 x 26,000

\Delta P=61,952.8\ lb/ft^2

So pressure difference will be \Delta P=61,952.8\ lb/ft^2.

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Calculate the reluctance of a 4-meter long toroidal coil made of low-carbon steel with an inner radius of 1.75 cm and an outer r
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Answer:

R = 31.9 x 10^(6) At/Wb

So option A is correct

Explanation:

Reluctance is obtained by dividing the length of the magnetic path L by the permeability times the cross-sectional area A

Thus; R = L/μA,

Now from the question,

L = 4m

r_1 = 1.75cm = 0.0175m

r_2 = 2.2cm = 0.022m

So Area will be A_2 - A_1

Thus = π(r_2)² - π(r_1)²

A = π(0.0225)² - π(0.0175)²

A = π[0.0002]

A = 6.28 x 10^(-4) m²

We are given that;

L = 4m

μ_steel = 2 x 10^(-4) Wb/At - m

Thus, reluctance is calculated as;

R = 4/(2 x 10^(-4) x 6.28x 10^(-4))

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3 years ago
What are the de Broglie frequencies and wavelengths of (a) an electron accelerated to 50 eV (b) a proton accelerated to 100 eV
DaniilM [7]

Answer:

(a) De-Brogie wavelength is 0.173 nm and frequency is 2.42 x 10^16 Hz

(b) De-Brogie wavelength is 2.875 pm and frequency is 4.8 x 10^16 Hz

Explanation:

(a)

First, we need to find velocity of electron. Since, it is accelerated by electric potential. Therefore,

K.E of electron = (1/2)mv² = (50 eV)(1.6 x 10^-19 J/1 eV)

(1/2)mv² = 8 x 10^(-18) J

Mass of electron = m = 9.1 x 10^(-31) kg

Therefore,

v² = [8 x 10^(-18) J](2)/(9.1 x 10^(-31) kg)

v = √1.75 x 10^13

v = 4.2 x 10^6 m/s

Now, the de Broglie's wavelength is given as:

λ = h/mv

where,

h = Plank's Constant = 6.626 x 10^(-34) kg.m²/s

Therefore,

λ = (6.626 x 10^(-34) kg.m²/s)/(9.1 x 10^(-31) kg)(4.2 x 10^6 m/s)

<u>λ = 0.173 x 10^(-9) m = 0.173 nm</u>

The frequency is given as:

Frequency = f = v/λ

f = (4.2 x 10^6 m/s)/(0.173 x 10^(-9) m)

<u>f = 2.42 x 10^16 Hz</u>

(b)

First, we need to find velocity of proton. Since, it is accelerated by electric potential. Therefore,

K.E of proton = (1/2)mv² = (100 eV)(1.6 x 10^-19 J/1 eV)

(1/2)mv² = 1.6 x 10^(-17) J

Mass of proton = m = 1.67 x 10^(-27) kg

Therefore,

v² = [1.6 x 10^(-17) J](2)/(1.67 x 10^(-27) kg)

v = √1.916 x 10^10

v = 1.38 x 10^5 m/s

Now, the de Broglie's wavelength is given as:

λ = h/mv

where,

h = Plank's Constant = 6.626 x 10^(-34) kg.m²/s

Therefore,

λ = (6.626 x 10^(-34) kg.m²/s)/(1.67 x 10^(-27) kg)(1.38 x 10^5 m/s)

<u>λ = 2.875 x 10^(-12) m = 2.875 pm</u>

The frequency is given as:

Frequency = f = v/λ

f = (1.38 x 10^5 m/s)/(2.875 x 10^(-12) m)

<u>f = 4.8 x 10^16 Hz</u>

6 0
3 years ago
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