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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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10) Um viajante, ao desembarcar no aeroporto de Londres, observou que o valor da temperatura do ambiente na escala Fahrenheit é
joja [24]

Answer:

The observed temperature was 10º Celsius or 50º Fahrenheit.

Explanation:

The traveler observed that the temperature in Fahrenheit is five times the value of the temperature in Celsius, therefore:

F = 5*C

A Fahrenheit temperature relates to a Celsius one by the following expression:

F = \frac{9}{5}*C + 32

Using the second expression on the first, we can solve for the temperature in Celsius, this is done below:

\frac{9}{5}*C + 32 = 5*C\\\frac{9}{5}*C - 5*C = -32\\\frac{9*C - 25*C}{5} = -32\\\frac{-16*C}{5} = -32\\-16*C = -32*5\\-16*C = -160\\C = \frac{-160}{-16} = 10\º\text{C}\\F = 5*C = 5*10 = 50\º\text{F}

The observed temperature was 10º Celsius or 50º Fahrenheit.

7 0
3 years ago
Please help me
bonufazy [111]

Переходи на сайте irkmix.top и получай много эмоций из Russia/

7 0
3 years ago
A Hooke's law spring is compressed 12.0 cm from equilibrium, and the potential energy stored is 72.0 J. What compression (as mea
Anuta_ua [19.1K]

Answer:14.14 cm

Explanation:

Given

Spring Compression x=12 cm

Potential energy Stored in spring=72 J

Suppose k is the spring constant of spring

Potential Energy of spring is given by =\frac{kx^2}{2}

\frac{k(0.12)^2}{2}=72

k(0.12)^2=144

k=10,000 N/m

k=10 kN/m

for 100 J energy

\frac{k(x_0)^2}{2}=100

10\times 10^3\cdot (x_0)^2=200

(x_0)^2=2\times 10^{-2}

x_0=0.1414

x_0=14.14 cm

6 0
3 years ago
the density of gold is 19.3 g/cm3. suppose a certain gold wedding ring deplaced 0.55mL of liquid when dropped in a glass of spar
adell [148]

Answer:

19.3

Explanation:

Assuming we have to find Specific gravity of gold.

As we know that specific gravity is defined as the ratio of weight of the object and weight of the water displaced by the object

so it is given by

specific gravity = weight of the object/weight of the water displaced

now we have

weight of the object = (density)(volume)g

weight of object = (19.3)(0.55)g

now weight of the liquid displaced is given by

weight of water displaced = (1 g/cm^3)(0.55ml)g

now we have

specific gravity = (19.3×0.55)/(1×0.55)

specific gravity= 19.3

8 0
3 years ago
A force gives a 5.0 kg object an acceleration of 2.0 m/s 2. The same force would give a 20 kg object an acceleration of _____. 0
Oksi-84 [34.3K]

m = 5 kg

a = 2 m/s²

to find the force that accelerates the 4 kg object @ 2 m/s²

F = ma = 5 kg x 2 m/s² = 10 N

To find what acceleration 10 N would give a 20 kg object

a = F/m = 10 N/20 kg = 0.5 m/s

6 0
3 years ago
Read 2 more answers
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