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Simora [160]
2 years ago
13

You are concerned about the moon roof on your new sports car. It seems to flex when driving at high speeds. Calculate how much n

et force the moon roof must withstand and in what direction. Assume the moon roof is flat with an area of 0.5 m^2 and the pressure and velocity over the moon roof is constant. Your driving speed is 20 m/s, and the velocity over the moon roof is 30 m/s. The pressure inside the car is 90,500 N/m^2 and the freestream pressure and density in front of the car are 90,000 N/m^2 and 1.1 kg/m^3.
Physics
1 answer:
Annette [7]2 years ago
4 0

Answer:

Explanation:

We shall apply Bernoulli's formula to solve the problem . It is as follows .

P + ρ gh + 1/2 ρ v² = constant .

P₁ + ρ gh + 1/2 ρ v₁²  = P₂ + ρ gh + 1/2 ρ v₂²

P₁ +  1/2 ρ v₁²  = P₂  + 1/2 ρ v₂²

P₁ - P₂ = 1/2 ρ (v₂² - v₁² )

= .5 x 1,1 ( 30² - 20² )

= 275 N / m²

velocity over moon roof is high , pressure will be lower there by 275 N / m²

Given pressure difference already existing = 90500 - 90000 = 500 N / m²

Additional pressure difference due to velocity difference = 275 N / m²

Total pressure difference = 275 + 500 = 775 N / m²

Area of roof = .5 m²

Total force acting upwards on the roof

= .5 x 775 N

= 387.5 N .

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Tina drove 3 miles north and 4 miles west to get to her grandmother’s house. Find the distance and displacement.​
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The area of land where water is drained downhill into a body of water is known as a drainage basin, or _______.
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Air pressure increases as you travel higher above sea level. This is the reason that cabins in commercial airliners require pres
irakobra [83]

The answer is true about the cabins in commercial airliners that require pressurization.

<h3>Why are the cabins of commercial airplanes pressurized?</h3>

Airplanes are pressurized because the air is very thin at the high altitude where they fly. The passenger jet has a cruising altitude of about 30,000 - 40,000 feet. At this altitude or height, humans can't breathe very well and our body gets less amount of oxygen. Most aircraft cabins are pressurized to an altitude about 8,000 feet. This is called cabin altitude. Aircraft pilots have access to the control's mode of a cabin pressure control system and if needed it can command the cabin to depressurize.

So we can conclude that cabins in commercial airliners require pressurization because of the greater pressure of the surrounding environment.

Learn more about pressure here: brainly.com/question/28012687

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4 0
10 months ago
A mass of 20 g stretches a spring 5 cm. Suppose that the mass is also attached to a viscous damper with a damping constant of 40
11111nata11111 [884]

Quasi frequency = 4√6

Quasi period = π√6/12

t ≈ 0.4045

<u>Explanation:</u>

Given:

Mass, m = 20g

τ = 400 dyn.s/cm

k = 3920

u(0) = 2

u'(0) = 0

General differential equation:

mu" + τu' + ku = 0

Replacing the variables with the known value:

20u" + 400u' + 3920u = 0

Divide each side by 20

u" + 20u' + 196u = 0

Determining the characteristic equation by replacing y" with r², y' with r and y with 1 in the differential equation.

r² + 20r + 196 = 0

Determining the roots:

r = \frac{-20 +- \sqrt{(20)^2 - 4(1)(196)} }{2(1)}

r = -10 ± 4√6i

The general solution for two complex roots are:

y = c₁ eᵃt cosbt + c₂ eᵃt sinbt

with a the real part of the roots and b be the imaginary part of the roots.

Since, a = -10 and b = 4√6

u(t) = c₁e⁻¹⁰^t cos 4√6t + c₂e⁻¹⁰^t sin 4√6t

u(0) = 2

u'(0) = 0

(b)

Quasi frequency:

μ = \frac{\sqrt{4km - y^2} }{2m}

= \frac{\sqrt{4(3929)(20) - (400)^2} }{2(20)} \\\\= 4\sqrt{6}

(c)

Quasi period:

T = 2π / μ

T = \frac{2\pi }{4\sqrt{6} } \\\\T = \frac{\pi\sqrt{6}  }{12}

(d)

|u(t)| < 0.05 cm

u(t) = |2e⁻¹⁰^t cos 4√6t + 5√6/6 e⁻¹⁰^t sin 4√6t < 0.05

solving for t:

τ = t ≈ 0.4045

8 0
2 years ago
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