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Nataly [62]
3 years ago
15

An airplane is flying horizontally with a constant momentum during a time interval ?t. (a) Is there a net impulse acting on the

plane during this time? Use the impulse-momentum theorem to guide your thinking. (b) In the horizontal direction, both the thrust generated by the engines and air resistance act on the plane. Considering your answer to part (a), how is the impulse of the thrust related in magnitude and direction to the impulse of the force due to the air resistance?
Physics
1 answer:
goldenfox [79]3 years ago
7 0

Answer:

Explanation:

Given

Airplane is flying with horizontally with a constant momentum during time interval \Delta t

Impulse is given by change in momentum, so there is no net impulse on the Plane because momentum is constant

(b) As there is no change in momentum therefore impulse of thrust and air drag is balanced i.e. both are equal in magnitude but act in opposite direction                            

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Answer:

1626.4 N

Explanation:

Given that a 82 kg man, at rest, drops from a diving board 3.0 m above the surface of the water and comes to rest 0.55 s after reaching the water. What force does the water exert on him?

The parameters to be considered are:

Distance S = 3m

Time t = 0.55s

Since the man started from rest, initial velocity u = 0

Using second equation of motion

S = Ut + 1/2at^2

3 = 1/2 × a × 0.55^2

3 = 1/2 × a × 0.3025

a = 3/ 0.15125

a = 19.83 m/s^2

Force = mass × acceleration

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3 years ago
All of the following are examples of suspensions except:
TEA [102]
Salt water is not an example of suspension as salt dissolves in water and combines with it rather than float in the water.
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What happens to the force between two charges if one of the charges are doubled?
Julli [10]

Answer:

The new force between the charges becomes double of the initial force.

Explanation:

The force acting between charge particles is given by :

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If one of the charges are doubled, then, q₁ = 2q₁

The new force becomes,

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7 0
4 years ago
Sound with a frequency of 1250 Hz leaves a room through a doorway with a width of 1.05 m.At what minimum angle relative to the c
kiruha [24]

Answer:

15.19°, 31.61°, 51.84°

Explanation:

We need to fin the angle for m=1,2,3

We know that the expression for wavelenght is,

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Substituting,

\lambda = \frac{344}{1250}

\lambda = 0.2752m

Once we have the wavelenght we can find the angle by the equation of the single slit difraction,

sin\theta = \frac{m \lambda}{W}

Where,

W is the width

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Re-arrange the expression,

\theta = sin^{-1} \frac{m\lambda}{W}

For m=1,

\theta = sin^{-1} \frac{1 (0.2752)}{1.05}= 15.19\°

For m=2,

\theta = sin^{-1} \frac{2 (0.2752)}{1.05}= 31.61\°

For m=3,

\theta = sin^{-1} \frac{3 (0.2752)}{1.05}= 51.84\°

<em>The angle of diffraction is directly proportional to the size of the wavelength.</em>

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