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ANTONII [103]
3 years ago
14

A goose with a mass of 2.0 kg strikes a commercial airliner with a mass of 160,000 kg head-on. Before the collision, the goose w

as flying with a speed of 60 km/hr and the aeroplane’s speed was 870 km/hour. Take the length of the goose to be 1.0 m long. (a) What is the change in momentum of the goose during this interaction?
Physics
1 answer:
Goryan [66]3 years ago
4 0

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

where;

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

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

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

v = 241.664 m/s

Thus, the final velocity of the bird is negligible compared to final  velocity of the airliner.

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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7

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Two friends leave a movie theater and take different busses to the same ice cream shop. one bus takes a longer route driving on
irinina [24]

The friend who has the greater displacement is the one who takes the longest path.

<h3>What is displacement?</h3>

The displacement is the shortest distance travelled by the particle. It is the vector quantity which re[presents both the magnitude and direction.

Given, two friends leave a movie theater and take different buses to the same ice cream shop. One bus takes a longer route driving on a high-speed highway, while the other takes a shorter route on lower-speed local roads.

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Thus, friend who has longer path has larger displacement.

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brainly.com/question/11934397

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8 0
2 years ago
What is the force needed to give a .25 kg arrow an acceleration of 196m/s2?
Rzqust [24]
<span>49N is the force needed to give a .25 kg arrow an acceleration of 196m/s2. F =ma ⇒ =( 0.25kg)(196m/s2) = 49N if the arrow is shot horizontally where the applied force is entirely in the x-direction.</span>
5 0
3 years ago
Read 2 more answers
A 1-kg rock is suspended by a massless string from one end of a
maxonik [38]

Answer:

The weight of measuring stick is 9.8 N

Explanation:

given information:

the mass of the rock, m_{r} = 1 kg

measuring stick, x =1 m

d = 0.25 m

to find the weight of measuring stick, we can use the following equation:

τ = Fd

τ = 0

F_{r} d - F_{s}d = 0

F_{r} = the force of the rock

F_{s} = the force of measuring stick

F_{s} =F_{r}

    = m g

    = 1 kg x 9.8 m/s

    = 9.8 N

thus, the weight of measuring stick is 9.8 N

6 0
3 years ago
Assume that the driver begins to brake the car when the distance to the wall is d=107m, and take the car's mass as m-1400kg, its
Evgen [1.6K]

Answer:

Explanation:

a ) Let let the frictional force needed be F

Work done by frictional force = kinetic energy of car

F x 107 = 1/2 x 1400 x 35²

F = 8014 N

b )

maximum possible static friction

= μ mg

where μ is coefficient of static friction

= .5 x 1400 x 9.8

= 6860 N

c )

work done by friction for μ = .4

= .4 x 1400 x 9.8 x 107

= 587216 J

Initial Kinetic energy

= .5 x 1400 x 35 x 35

= 857500 J

Kinetic energy at the at of collision

= 857500 - 587216

= 270284 J

So , if v be the velocity at the time of collision

1/2 mv² = 270284

v = 19.65 m /s

d ) centripetal force required

= mv₀² / d which will be provided by frictional force

= (1400 x 35 x 35) / 107

= 16028 N

Maximum frictional force possible

= μmg

= .5 x 1400 x 9.8

= 6860 N

So this is not possible.

4 0
3 years ago
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