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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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Trava [24]

In a spark ignition engine, the fuel is mixed with air and then inducted into the cylinder during the intake process. After the piston compresses the fuel-air mixture, the spark ignites it, causing combustion. The expansion of the combustion gases pushes the piston during the power stroke.

Credit: Department of Energy

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2 years ago
You drop a steel ball bearing, with a radius of 2.40 mm, into a beaker of honey. Note that honey has a viscosity of 6.00 Pa/s an
Stells [14]

Answer:

The “terminal speed” of the ball bearing is 5.609 m/s

Explanation:

Radius of the steel ball R = 2.40 mm

Viscosity of honey η = 6.0 Pa/s

\text { Viscosity has Density } \sigma=1360 \mathrm{kg} / \mathrm{m}^{3}

\text { Steel has a density } \rho=7800 \mathrm{kg} / \mathrm{m}^{3}

\left.\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2} \text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^{2} \text { on Earth }\right)

While calculating the terminal speed in liquids where density is high the stokes law is used for viscous force and buoyant force is taken into consideration for effective weight of the object. So the expression for terminal speed (Vt)

V_{t}=\frac{2 \mathrm{R}^{2}(\rho-\sigma) \mathrm{g}}{9 \eta}

Substitute the given values to find "terminal speed"

\mathrm{V}_{\mathrm{t}}=\frac{2 \times 0.0024^{2}(7800-1360) 9.8}{9 \times 6}

\mathrm{V}_{\mathrm{t}}=\frac{0.0048 \times 6440 \times 9.8}{54}

\mathrm{V}_{\mathrm{t}}=\frac{302.9376}{54}

\mathrm{V}_{\mathrm{t}}=5.609 \mathrm{m} / \mathrm{s}

The “terminal speed” of the ball bearing is 5.609 m/s

7 0
3 years ago
A convex lens has a focal length of 16.5 cm. Where on the lens axis should an object be placed in order to get a virtual, enlarg
alexandr402 [8]

Answer:

Object should be placed at a distance, u = 7.8 cm

Given:

focal length of convex lens, F = 16.5 cm

magnification, m = 1.90

Solution:

Magnification of lens, m = -\frac{v}{u}

where

u = object distance

v = image distance

Now,

1.90 = \frac{v}{u}

v = - 1.90u

To calculate the object distance, u by lens maker formula given by:

\frac{1}{F} = \frac{1}{u}+ \frac{1}{v}

\frac{1}{16.5} = \frac{1}{u}+ \frac{1}{- 1.90u}

\frac{1}{16.5} = \frac{1.90 - 1}{1.90u}

\frac{1}{16.5} = \frac{ 0.90}{1.90u}

u = 7.8 cm

Object should be placed at a distance of 7.8 cm on the axis of the lens to get virtual and enlarged image.

6 0
3 years ago
A stunt woman attempts to swing from the roof of a m 24 high building to the bottom of an identical building using a m 24 rope,
MakcuM [25]

Answer:

Height from ground is 8 m where string will break

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Let the string makes some angle with the vertical after some instant of time

So here we have

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now by energy conservation we have

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T = mg cos\theta + 2mgcos\theta

T = 3mg cos\theta

For string break down we have

T = 2mg = 3mgcos\theta

cos\theta = \frac{2}{3}

Now height from the ground is given as

h = R - Rcos\theta

h = 24 - 24(\frac{2}{3})

h = 8 cm

7 0
3 years ago
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3 years ago
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