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lesya [120]
3 years ago
9

Canada geese migrate essentially along a north–south direction for well over a thousand kilo-meters in some cases, traveling at

speeds up to about 100 km/h. If one goose is flying at 100 km/h relative to the air but a 40-km/h wind is blowing from west to east, (a) at what angle relative to the north–south direction should this bird head to travel directly southward relative to the ground? (b) How long will it take the goose to cover a ground distance of 500 km from north to south? (Note: Even on cloudy nights, many birds can navigate by using the earth’s magnetic field to fix the north–south direction.)
Physics
1 answer:
kotykmax [81]3 years ago
7 0

Answer:

a) 66.4 relative to the west in the south-west direction

b) 5.455 hours

Explanation:

a)If the wind is blowing east-ward at a speed of 40km/h, then the west component of the geese velocity must be 40km/h in order to counter balance it. Geese should be flying south-west at an angle of

cos(\alpha) = 40 / 100 = 0.4

\alpha = cos^{-1}(0.4) = 1.16 rad = 180\frac{1.16}{\pi} = 66.4^0 relative to the West

b) The south-component of the geese velocity is

100sin(\alpha) = 100sin(66.4^0) = 91.65 km/h

The time it would take for the geese to cover 500km at this rate is

t = 500 / 91.65 = 5.455 hours

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A steady beam of alpha particles (q = + 2e, mass m = 6.68 × 10-27 kg) traveling with constant kinetic energy 22 MeV carries a cu
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Answer:

Explanation:

q = 2e = 3.2 x 10^-19 C

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Current, i = 0.27 micro Ampere = 0.27 x 10^-6 A

(a) time, t = 2.8 s

Let N be the alpha particles strike the surface.

N x 2e = q

N x 3.2 x 10^-19 = i t

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N = 2.36 x 10^12

(b) Length, L = 16 cm = 0.16 m

Let N be the alpha particles

K = 0.5 x mv²

22 x 1.6 x 10^-13 = 0.5 x 6.68 x 10^-27 x v²

v² = 1.054 x 10^15

v = 3.25 x 10^7 m/s

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22 x 1.6 x 10^-13 = 2 x 1.5 x 10^-19 x V

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From the question,

Given: q = 0.002 mC = 2.0×10⁻⁶ C, q' = 0.03 nC = 3.0×10⁻¹¹ C, r = 0.5 m

Constant: k = 8.99×10⁹ Nm²/C²

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