Answer:
The average momentum of the bird is 0.26 kgm/s
Explanation:
The formula to be used here is that of momentum which is
momentum (in kgm/s) = mass (in kg) × velocity (in m/s)
The velocity of the bird is
velocity (in m/s) = distance (in meter) ÷ time (in seconds)
distance in meters = 125km × 1000 = 125,000 m
time in seconds = 4 hrs × 60 × 60 = 14,400 secs
velocity = 125000/14400
velocity = 8.68 m/s
momentum (p) = 0.03 × 8.68
p = 0.26 kgm/s
The average momentum of the bird is 0.26 kgm/s
The average momentum of the bird (in unit vector notation) is (0.1842i + 0.1842j) kgm/s.
Since the unladen swallow that weighs 0.03 kg flies straight northeast (that is at a bearing of 45°) a distance of 125 km in 4.0 hours.
Its position vector after 4.0 hours is d = (125kmcos45)i + (125kmsin45)j = (125000 × 1/√2)i + (125000 × 1/√2)j
= (62500√2)i + (62500√2)j.
If the initial position of the swallow is d' = 0i + 0j, then its total displacement after 4 hours is, D = d - d'
= (62500√2)i + (625000√2)j - (0i + 0j)
= (62500√2)i + (62500√2)j m
The unladen swallow's average velocity, v = D/t where
So, v = [(62500√2)i + (62500√2)j m]/14400 s = (88388.35)i/14400 + (88388.35)j /1440
= 6.14i + 6.14j m/s
The average momentum of the unladen swallow is p = mv where
So, p = mv
p = 0.03 kg × (6.14i + 6.14j m/s)
p = (0.1842i + 0.1842j) kgm/s
So, the average momentum of the bird (in unit vector notation) is (0.1842i + 0.1842j) kgm/s.
Learn more about average momentum here:
https://brainly.com/question/25941500
