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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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Ratling [72]

Answer:

Explanation:

Given that, current generated from lightning range from

10⁴ A < I < 10^5 A

We know that,

The magnetic force is given as

F = iLB

The magnetic field on the earth surface is

B = 10^-5 T

So, let assume the worst case of a 15m flag pole

L = 15m

Then,

F = iLB

F = 10^5 × 10 × 10^-5

F = 15 N

Therefore, 15N is fairly strong so it will come to the material that was use for the material of the flag pole.

Therefore, it is possible that the student is right depending on the material of the flag pole.

7 0
3 years ago
Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface.
elena-s [515]

Answer:

g=13.42\frac{m}{s^2}

Explanation:

1) Notation and info given

\rho_{center}=13000 \frac{kg}{m^3} represent the density at the center of the planet

\rho_{surface}=2100 \frac{kg}{m^3} represent the densisty at the surface of the planet

r represent the radius

r_{earth}=6.371x10^{6}m represent the radius of the Earth

2) Solution to the problem

So we can use a model to describe the density as function of  the radius

r=0, \rho(0)=\rho_{center}=13000 \frac{kg}{m^3}

r=6.371x10^{6}m, \rho(6.371x10^{6}m)=\rho_{surface}=2100 \frac{kg}{m^3}

So we can create a linear model in the for y=b+mx, where the intercept b=\rho_{center}=13000 \frac{kg}{m^3} and the slope would be given by m=\frac{y_2-y_1}{x_2-x_1}=\frac{\rho_{surface}-\rho_{center}}{r_{earth}-0}

So then our linear model would be

\rho (r)=\rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r

Since the goal for the problem is find the gravitational acceleration we need to begin finding the total mass of the planet, and for this we can use a finite element and spherical coordinates. The volume for the differential element would be dV=r^2 sin\theta d\phi d\theta dr.

And the total mass would be given by the following integral

M=\int \rho (r) dV

Replacing dV we have the following result:

M=\int_{0}^{2\pi}d\phi \int_{0}^{\pi}sin\theta d\theta \int_{0}^{r_{earth}}(r^2 \rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r)

We can solve the integrals one by one and the final result would be the following

M=4\pi(\frac{r^3_{earth}\rho_{center}}{3}+\frac{r^4_{earth}}{4} \frac{\rho_{surface}-\rho_{center}}{r_{earth}})

Simplyfind this last expression we have:

M=\frac{4\pi\rho_{center}r^3_{earth}}{3}+\pi r^3_{earth}(\rho_{surface}-\rho_{center})

M=\pi r^3_{earth}(\frac{4}{3}\rho_{center}+\rho_{surface}-\rho_{center})

M=\pi r^3_{earth}[\rho_{surface}+\frac{1}{3}\rho_{center}]

And replacing the values we got:

M=\pi (6.371x10^{6}m)^2(\frac{1}{3}13000 \frac{kg}{m^3}+2100 \frac{kg}{m^3})=8.204x10^{24}kg

And now that for any shape the gravitational acceleration is given by:

g=\frac{MG}{r^2_{earth}}=\frac{(6.67408x10^{-11}\frac{m^3}{kgs^2})*8.204x10^{24}kg}{(6371000m)^2}=13.48\frac{m}{s^2}

4 0
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How ocean currents effect temperature and the amount of moisture of the air mass above coastlines
GarryVolchara [31]
Hotter ocean tempatures mean more moisture in the dense air mass
8 0
3 years ago
The average annual rainfall in Naples, Italy, is 39.6 inches, and the average annual rainfall in Fort Collins, Colorado, is 15 i
Nezavi [6.7K]
The simple answer would be; closeness to the equator combined with general climate.

7 0
3 years ago
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Part of one row of the periodic table is shown. 4 blue boxes in a row. The first has S n at the center and 50 above with tin and
nydimaria [60]

Answer:

Atoms of tellurium (Te) have the greatest average number of neutrons equal to 76.

Explanation:

In the periodic table, Elements are represented with their respected symbols. Above the symbol is the elements atomic number which is equal to the number of protons in each atom. Below the symbol is the mass number of that element which is roughly equal to the sum of neutrons and protons of that atom.

To calculate the number of neutrons we can take the difference of Atomic number and mass number:

Number of neutrons = mass number - atomic number

<u>- Tin:</u>

Atomic number = 50

Mass number = 119

Number of neutrons = mass number - atomic number = 119 - 50

Number of neutrons = 69

<u>- Antimony(Sb):</u>

Atomic number = 51

Mass number = 122

Number of neutrons = mass number - atomic number = 122 - 51

Number of neutrons = 71

<u>- Tellurium(Te):</u>

Atomic number = 52

Mass number = 128

Number of neutrons = mass number - atomic number = 128 - 52

Number of neutrons = <u>76</u>

<u>- Iodine(I):</u>

Atomic number = 53

Mass number = 127

Number of neutrons = mass number - atomic number = 127 - 53

Number of neutrons = 74

Here, the greatest number of neutrons is for the atoms of Tellurium(Te).

7 0
3 years ago
Read 2 more answers
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