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Airida [17]
3 years ago
13

you travel 4.0km east, 4.0km north, then 5.0km at 53.1 degrees north of west in a total of 5 hours. What is the magnitude and di

rection of your average velocity
Physics
1 answer:
Sergio039 [100]3 years ago
7 0

Answer:

Explanation:

We shall convert all the displacement in vector form .

i and j represents east and north respectively .

D₁ = 4 i

D₂ = 4 j

D₃ = - 5 cos 53.1 i + 5 sin 53.1 j

= -3i + 4 j

Total displacement = D₁ + D₂ + D₃

= 4i + 4 j - 3i + 4 j

= i + 8j

magnitude of displacement = √( 1² + 8² )

= 8.06 km

velocity = 8.06 / 5

= 1.61 km / h

Direction from x axis in anticlockwise direction .

Tanθ =  8 / 1 = 8

θ = 83° north of east .

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Answer:

The change in momentum of the goose during this interaction is 33.334 m/s

Explanation:

Given;

mass of goose, m₁ = 2.0 kg

mass of commercial airliner, m₂ = 160,000 kg

initial velocity of the bird, u₁ = 60 km/hr  = 16.667 m/s

initial velocity of the airliner, u₂ = 870 km/hr = 241.667 m/s

Change in momentum is given as;

ΔP = mv - mu

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u is the initial velocity of the bird

v is the final velocity of the bird

Apply the principle of conservation of linear momentum;

Total momentum before collision = Total momentum after collision

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v is the final velocity of bird and airliner after collision;

(2 x 16.667) + (160,000 x 241.667) = v (2 + 160,000)

38,666,753.334 = 160,002v

v = 38,666,753.334 / 160,002

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ΔP = mv - mu

ΔP = m(v - u)

ΔP = 2(0 - 16.667)

ΔP = -33.334 m/s

The negative sign implies a deceleration of the bird after the impact.

Therefore, the change in momentum of the goose during this interaction is 33.334 m/s

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