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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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The figure above shows the net force exerted on an object as a function of the position of the object. The object starts from re
weqwewe [10]

Answer:

0.06 Kg

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Final velocity (v) = 3.0 m/s

Distance (s) = 0.09 m

Net Force (F) = 3 N

Mass (m) =?

Next, we shall determine the acceleration of the object. This can be obtained as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 3.0 m/s

Distance (s) = 0.09 m

Acceleration (a) =?

v² = u² + 2as

3² = 0² + (2 × a × 0.09)

9 = 0 + 0.18a

9 = 0.18a

Divide both side by 0.18

a = 9 / 0.18

a = 50 m/s²

Finally, we shall determine the mass of the object. This can be obtained as follow:

Net Force (F) = 3 N

Acceleration (a) = 50 N

Mass (m) =?

F = ma

3 = m × 50

Divide both side by 50

m = 3 / 50

m = 0.06 Kg

Therefore, the mass of the object is 0.06 Kg

5 0
2 years ago
One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a
Maurinko [17]

This question is not complete.

The complete question is as follows:

One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a space station that spins about its center at a constant rate. This creates “artificial gravity” at the outside rim of the station. (a) If the diameter of the space station is 800 m, how many revolutions per minute are needed for the “artificial gravity” acceleration to be 9.80m/s2?

Explanation:

a. Using the expression;

T = 2π√R/g

where R = radius of the space = diameter/2

R = 800/2 = 400m

g= acceleration due to gravity = 9.8m/s^2

1/T = number of revolutions per second

T = 2π√R/g

T = 2 x 3.14 x √400/9.8

T = 6.28 x 6.39 = 40.13

1/T = 1/40.13 = 0.025 x 60 = 1.5 revolution/minute

6 0
3 years ago
A speedboat initially at rest accelerates at 4.0 m/s (squared) for 7.0 s. How far does the speedboat move in 7.0 s?
Ulleksa [173]
I think is 3 becuase when he star at 4
4 0
2 years ago
What is the energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b?
xeze [42]

The energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b is 3.06 eV.

<h3>Change in energy level of the electron</h3>

When photons jump from a higher energy level to a lower level, they emit or radiate energy.

The change in energy level of the electrons is calculated as follows;

ΔE = Eb - Ef

ΔE = -2.68 eV - (-5.74 eV)

ΔE = 3.06 eV

Thus, the energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b is 3.06 eV.

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7 0
2 years ago
all objects near the earths surface-regardless of size and weight have the same force of gravity acting on them. is it true or f
pickupchik [31]

False: the force of gravity acting on different objects is different and depends on their mass

Explanation:

The answer is false.

The force of gravity acting on an object (also known as weight) near the Earth's surface is given by:

F=mg

where:

m is the mass of the object

g=9.8 m/s^2 is the acceleration of gravity

We see from the formula that the force of gravity acting on an object depends on the mass: the larger the mass of the object, the stronger the gravitational force acting on it, and the smaller the mass, the weaker the force of gravity.

The factor that does not change is the acceleration of gravity, which is constant (9.8 m/s^2) if we are near the Earth's surface, and implies that all the objects in free fall accelerate at the same rate towards the ground, regardless of their size and weight.

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brainly.com/question/12978926

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5 0
3 years ago
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