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Ede4ka [16]
3 years ago
6

An aircraft flies at sea level at a speed of 220 m/s. What is the highest pressure that can be acting on the surface of the airc

raft?
Physics
2 answers:
Nadusha1986 [10]3 years ago
7 0

Answer:

highest pressure = 101325 + 29524 = 130849 Pa

Explanation:

the highest pressure that can act at the sea level is the sum of dynamic pressure (due to flowing fluid) and the static pressure (due to stationary fluid).

P =P_static + P-dynamic

P_static = 1 atm = 101325 Pa

P_dynamic = 0.5*p*v^2

where p: density of fluid = 1.22 kg/m^3 at sea level

           v: fluid velocity = 220m/s

P_dynamic = 0.5 * 1.22*220^2 = 29524 Pa

highest pressure = 101325 + 29524 = 130849 Pa

goldenfox [79]3 years ago
6 0

Answer:

An aircraft flying at sea level with a speed of 220 m/s, has a highest pressure of 29136.8 N/m²

Explanation:

Applying Bernoulli's equation, we determine the highest pressure on the aircraft.

P = \frac{1}{2} \rho V^2

where;

P is the highest pressure on the aircraft

\rho is the density of air = 1.204 kg/m³ at sea level temperature.

V is the velocity of the aircraft = 220 m/s

P = 0.5*1.204*(220)² = 29136.8 N/m²

Therefore, an aircraft flying at sea level with a speed of 220 m/s, has a highest pressure of 29136.8 N/m²

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6 beats

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3 years ago
A turntable that spins at a constant 80.0 rpmrpm takes 3.50 ss to reach this angular speed after it is turned on. Find its angul
Veronika [31]

Answer:

The angular acceleration is <u>2.39 rad/s²</u>.

The number of degrees it rotates is <u>841.68 degrees</u>.

Explanation:

Given:

Initial angular speed (ω₀) = 0 rad/s

Final angular speed in rpm (N) = 80.0 rpm

Time taken (t) = 3.50 s

First, let us determine the final angular speed in radians per second.

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\omega=\frac{2\pi N}{60}\ rad/s

Plug in the values and find the final angular speed, 'ω'. This gives,

\omega=\frac{2\pi\times 80.0}{60}=8.38\ rad/s

Now, using equation of motion for rotational motion, we have:

\omega=\omega_0+\alpha t\\\\\alpha\to angular\ acceleration

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8.38=0+\alpha \times 3.50\\\\\alpha=\frac{8.38}{3.50}=2.39\ rad/s^2

Therefore, the angular acceleration is 2.39 rad/s².

Now, again using rotational equation of motion relating angular displacement, we have:

\omega^2=\omega_0^2+2\alpha\theta\\\\\theta=\frac{\omega^2-\omega_0^2}{2\alpha }

Plug in the given values and solve for 'θ'. This gives,

\theta=\frac{(8.38)^2-0}{2\times 2.39}\\\\\theta=\frac{70.2244}{4.78}=14.69\ rad

Convert radians to degrees using the conversion factor. This gives,

π radians = 180°

So, 1 radian =( 180 ÷ π ) degrees

Therefore, 14.69\ rad=14.69\times (\frac{180}{\pi})=841.68^\circ

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3 0
3 years ago
An object attached to an ideal spring oscillates with an angular frequency of 2.81 rad/s. the object has a maximum displacement
NikAS [45]
Ω = 2.81
A = 0.232
k = 29.8

x = A cos(ωt + Ф)

at t = 0:
x = A = A cos(ωt + Ф) = A cos(Ф)
Ф = 0

at t = 1.42, with Ф = 0:
x = A cos(ωt) 

U = 1/2 k x² = 1/2 k [A cos(ωt)]²
4 0
3 years ago
8.92 A 45.0 kg woman stands up in a 60.0 kg canoe 5.00 m long. She walks from a point 1.00 m from one end to a point I .00 m fro
omeli [17]

The distance that the canoe moves in this process is 1.29 meters.

We first have to find the center of mass

X = \frac{MsXs+McXc}{Mw+Mc} \\\\

Where

Ms = Woman's mass = 45

Mc = Canoe's mass = 60kg

Xs = position from left= 1 cm

Xc = position from left end of canoe's mass = 2.5cm

When we put these values into the equation we have:

X=\frac{45*1+60*2.5}{45+60} \\\\= 1.857\\

The center of gravity lies at the center of this boat. Therefore,

Xc = \frac{L}{2} \\\\L = 5 m long\\\\\frac{5}{2} =2.5

5.00 - 1. 00 = 4 meters

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To get the distance that is moved by this canoe

distance = 3.143-1.857

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The distance that the canoe moves in this process is 1.29 meters.

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8-30-2018 what phase change occurs as the substance at 0 degree c has its pressure dropped from 0.50 at to 0.25 atm
madam [21]

Answer:

at T = 0ºC the change of state is from the solid state to the gaseous state

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