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Ede4ka [16]
2 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]2 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]2 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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A car is traveling at a 20.0 m/s for 7.00 s and then suddenly comes to a stop over a 3 s period.
Delicious77 [7]

Answer: A) Deceleration of the car is -6.6667 m/s² while it came to stop.

B) The total distance the car travels is 200 meter during the 10 s period.

Explanation:

Given Data

Initial velocity of the car ($$v_{i}$$) =   20.0 m/s

Final velocity of the car (v_{f}) = 0 m/s

Time (in motion) =7.00 s

Time (in rest) =3 s

To find - A) car's deceleration while it came to a stop

              B) the total distance the car travels in 10 s

A) The formula to find the deceleration is

Deceleration = (( final velocity - initial velocity ) ÷ Time)        (m/s²)

Deceleration = ((v_{f}) - ($$v_{i}$$)) ÷ time     (m/s²)

Deceleration =  ( 0.0 - 20 ) ÷ 3      (m/s²)

Deceleration =   (- 20) ÷ 3  (m/s²)

Deceleration   =  - 6.6667 m/s²

(NOTE : Deceleration is the opposite of acceleration so the final result must have the negative sign)

The car's deceleration is  - 6.6667 m/s² while it came to a stop

B) The formula to find the distance traveled by the car is  

Distance traveled by the car is equals to the product of the speed and time

Distance = Speed × Time  (meter)

Distance = 20.0 × 10

Distance = 200 meters

The total distance the car travels during the period of 10 s is 200 meters

7 0
2 years ago
A 3.00-kg object undergoes an acceleration given by a = (2.00 i + 5.00 j) m/s^2. Find (a) the resultant force acting on the obje
kobusy [5.1K]

Answer:

(a): The resultant force acting on the object are F= (5.99 i + 14.98 j).

(b): The magnitude of the resultant force are F= 16.4 N < 68.19º .

Explanation:

m= 3kg

a= 2 i + 5 j = 5 .38 < 68.19 º

F= m * a

F= 3* ( 5.38 < 68.19º )

F= 16.4 N < 68.19º

Fx= F * cos(68.19º)

Fx= 5.99

Fy= F* sin(68.19º)

Fy= 14.98

3 0
3 years ago
I will be so thankful if u answer correctly!!​
olga_2 [115]
<h2>Answer:</h2>

(a) 10N

<h2>Explanation:</h2>

The sketch of the two cases has been attached to this response.

<em>Case 1: The box is pushed by a horizontal force F making it to move with constant velocity.</em>

In this case, a frictional force F_{r} is opposing the movement of the box. As shown in the diagram, it can be deduced from Newton's law of motion that;

∑F = ma    -------------------(i)

Where;

∑F = effective force acting on the object (box)

m = mass of the object

a = acceleration of the object

∑F = F -  F_{r}

m = 50kg

a = 0   [At constant velocity, acceleration is zero]

<em>Substitute these values into equation (i) as follows;</em>

F -  F_{r} = m x a

F -  F_{r} = 50 x 0

F -  F_{r} = 0

F =  F_{r}            -------------------(ii)

<em>Case 2: The box is pushed by a horizontal force 1.5F making it to move with a constant velocity of 0.1m/s²</em>

In this case, the same frictional force F_{r} is opposing the movement of the box.

∑F = 1.5F -  F_{r}

m = 50kg

a =  0.1m/s²

<em>Substitute these values into equation (i) as follows;</em>

1.5F -  F_{r} = m x a

1.5F -  F_{r} = 50 x 0.1

1.5F -  F_{r} = 5            ---------------------(iii)

<em>Substitute </em>F_{r}<em> = F from equation (ii) into equation (iii) as follows;</em>

1.5F - F = 5            

0.5F = 5            

F = 5 / 0.5

F = 10N

Therefore, the value of F is 10N

<em />

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2 years ago
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frez [133]

Explanation:

It represents the direction of flow of positive charge but is treated as a scalar quantity because current follows the laws of scalar addition and not the laws of vector addition. The angle between the wires carrying current does not affect the total current in the circuit.

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No,because they  may have more particles 
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