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Lorico [155]
3 years ago
15

At takeoff, an aircraft travels at 62 m/s, so that the air speed relative to the bottom of the wing is 62 m/s. Given the sea lev

el density of air to be 1.29 kg/m3, how fast must it move over the upper surface to create the ideal lift?
Physics
1 answer:
olganol [36]3 years ago
5 0

Answer:

the aircraft must travel at a speed of <em>73.4 m/s</em> in order to create the ideal lift.

Explanation:

We will use Bernoulli's theorem in order to determine the pressure lift:

ΔP = 1/2 (ρ)(v₂² - v₁²)

the generated pressure lift is ΔP = 1000 N/m²

Therefore,

1000 = 1/2(ρ)(v₂² - v₁²)

v₂² - v₁² = 2000 / ρ

v₂² = (2000 N/m² / 1.29 kg/m³) + (62 m/s)²

v₂ = √[ (2000 N/m² / 1.29 kg/m³) + (62 m/s)² ]

<em>v₂ = 73.4 m/s </em>

<em></em>

Therefore, the aircraft must travel at a speed of <em>73.4 m/s</em> in order to create the ideal lift.

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Answer:

7.4 watt

Explanation:

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Which quantity is a vector quantity A. Acceleration B. Mass C. Speed D. Volume
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Answer:

Acceleration

Explanation:

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3 years ago
It continues to fly along the same horizontal arc but increases its speed at the rate of 1.34 m/s 2 . Find the magnitude of acce
shepuryov [24]

This Question is not complete

Complete Question:

a. A hawk flies in a horizontal arc of radius 11.3 m at a constant speed of 5.7 m/s. Find its centripetal acceleration.

Answer in units of m/s2

b. It continues to fly along the same horizontal arc but increases its speed at the rate of 1.34 m/s2. Find the magnitude of acceleration under

these new conditions.

Answer in units of m/s2

Answer:

a. 2.875m/s²

b. 3.172m/s²

Explanation:

a. The formula for centripetal acceleration = (speed²) ÷ radius

Centripetal acceleration = (5.7m/s)²÷ 11.3m

Centripetal acceleration = 2.875m/s²

b. Magnitude of acceleration can be calculated by finding the sum of the vectors for the both the centripetal acceleration and the increase in the speed rate.

Centripetal acceleration ( acceleration x) = 2.875m/s²

Increase in the speed rate ( acceleration n) = 1.34m/s²

Magnitude of acceleration = √a²ₓ + a²ₙ

=√( 2.875m/s²)²+ (1.34m/s²)²

= √ 10.06m/s²

= 3.172m/s²

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3 years ago
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Answer:

49.3 N

Explanation:

Given that Pulling up on a rope, you lift a 4.25 kg bucket of water from a well with an acceleration of 1.80 m/s2 . What is the tension in the rope?

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Weight = 4.25 × 9.8

Weight = 41.65 N

The tension and the weight will be opposite in direction.

Total force = ma

T - mg = ma

Make tension T the subject of formula

T = ma + mg

T = m ( a + g )

Substitutes all the parameters into the formula

T = 4.25 ( 1.8 + 9.8 )

T = 4.25 ( 11.6 )

T = 49.3 N

Therefore, the tension in the rope is 49.3 N approximately.

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