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3 years ago

A light plane must reach a speed of 33 m/s per take off. How long a runway is needed if the constant acceleration is 3.0 m/s^2.

1 answer:

3 0

Given the final velocity (Vf) and the acceleration (a), the distance that should be traveled by the plane is calculated through the equation,
d = (Vf² - Vi²) / 2a
V1 should be zero because the light plane started the motion from rest. Substituting the given values,
d = ((33 m/s)² - 0)) / 2(3 m/s²)
The distance is therefore equal to 181.5 meters.

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The correct answer is the last one:
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The direction of an electric field is the direction (5 points)

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3 years ago

One of the most studied objects in the night sky is the Crab nebula, the remains of a supernova explosion observed by the Chines

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Answer:

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Explanation:

Given that,

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3 years ago

A horizontal force of magnitude 46.3 n pushes a block of mass 4.14 kg across a floor where the coefficient of kinetic friction i

IrinaVladis [17]

A) Calling F the intensity of the horizontal force and d the displacement of the block across the floor, the work done by the horizontal force is equal to
$W=Fd = (46.3 N)(4.25 m)=196.8 J$

b) The work done by the frictional force against the motion of the block is equal to:
$W_f = -F_f d =- (\mu mg) d =-(0.609)(4.14 kg)(9.81 m/s^2)(4.25 m)=$
$=-105.1 J$
Part of these 105.1 Joules of work becomes increase of thermal energy of the block ( $\Delta E_B$ ), and part of it becomes increase of thermal energy of the floor ( $\Delta E_F$ ). We already know the increase in thermal energy of the block (38.2 J), so we can find the increase in thermal energy of the floor:
$\Delta E_F = 105.1 J - \Delta E_B = 105.1 J-38.2 J=66.9 J$

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3 years ago

A 1500-kilogram car is at rest on a level track. The driver gives the engine gas and the car begins to accelerate forward. The c

Nutka1998 [239]

1. The acceleration of the car is 3.375 m/s².

2. The force moving the car is 5062.5 N

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

Mass of car = 1500 Kg

Initial velocity (u) = 0 m/s

Final velocity (v) = 27 m/s

Time (t) = 8 s

<h3>Acceleration (a) =? </h3><h3>Force (F) =? </h3>

<h3>1. Determination of the acceleration </h3>

Initial velocity (u) = 0 m/s

Final velocity (v) = 27 m/s

Time (t) = 8 s

<h3>Acceleration (a) =?</h3>

$a = \frac{v - u}{t} \\\\a = \frac{27 - 0}{8} \\\\a = \frac{27}{8}\\$

Therefore, the acceleration of the car is 3.375 m/s²

<h3>2. Determination of the force moving the car.</h3>

Mass of car = 1500 Kg

Acceleration (a) = 3.375 m/s²

<h3>Force (F) =?</h3>

F = 1500 × 3.375

Therefore, the force moving the car is 5062.5 N

Learn more: brainly.com/question/13001406

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3 years ago