1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Anestetic [448]
3 years ago
6

To check this out, calculate the induced voltage for a wild goose with a wingspan of 1.5 m at level flight due south at 13 m/s a

t a point where the earth's magnetic field is 3 ×10−5 T directed downward from horizontal by 14 degrees. What would be the expected voltage difference across the goose's wingtips?
Physics
1 answer:
lara [203]3 years ago
6 0

Complete question:

It is known that birds can detect the earth's magnetic field, but the mechanism of how they do this is not known. It has been suggested that perhaps they detect a motional EMF as they fly north to south, but it turns out that the induced voltages are small compared to the voltages normally encountered in cells, so this is probably not the mechanism involved.

To check this out, calculate the induced voltage for a wild goose with a wingspan of 1.5 m at level flight due south at 13 m/s at a point where the earth's magnetic field is 3 ×10−5 T directed downward from horizontal by 14 degrees. What would be the expected voltage difference across the goose's wingtips?

Answer:

The expected voltage difference across the goose's wingtips is 0.1415 mV

Explanation:

Given;

wingspan of the wild goose, L = 1.5 m

speed of the wild goose, v = 13 m/s

Earth's magnetic field strength, B =  3 × 10⁻⁵ T

angle of inclination of the bird's motion to Earth's magnetic field, θ = 14⁰

If the bird's wings act like conduct in the Earth's magnetic field, then induced emf according to Faraday, will be calculated as follows;

emf = BLvsinθ

where;

B is Earth's magnetic field strength

L is length of the conductor (wingspan)

v is velocity of the conductor (wingspan)

θ is the angle of inclination to the magnetic field

Induced emf = (3 × 10⁻⁵ )( 1.5 )(13 )sin14⁰

Induced emf = 1.4152 x 10⁻⁴ V = 0.1415 mV

Thus, the expected voltage difference across the goose's wingtips is 0.1415 mV

You might be interested in
How fast would you be going (in kmh) if you had a ship that accelerated at a constant 1g for 24 hours?
Nady [450]

Answer:

Explanation:

1 g is 9.8 m/s^2 the problem wants the results in km/h so we'll fix that really quick.

9.8 m/s^2 (1 km/1000m)(60 sec/1 min)^2(60 min/1 hour)^2 = 127008 km/hour^2

Now, I'm assuming the ship is starting from rest, and hopefully you know your physics equations.  We are going to use vf = vi + at.  Everything is just given, or we can assume, so I'll just solve.

vf = vi + at

vf = 0 + 127008 km/hour^2 * 24 hours

vf = 3,048,192 km/hour

If there's anything that doesn't make sense let me know.  

5 0
4 years ago
Pls help quickly
emmasim [6.3K]

Answer: frosted glass

Explanation:

8 0
3 years ago
Read 2 more answers
The form of energy that can move from place to place across the universe is . On Earth, the main source of this energy is .
serg [7]

The form of energy that can move from place to place across the universe is light energy. On earth, the main source of this energy is Sun. Most of the light energy comes from the sun because it is the primary source of all the energies. The food, fossil fuels, movement of winds, etc all exists due to Sun. Without sun, there won't be any light energy on the earth. In all the processes which occur on earth has a direct or indirect involvement of light energy which comes from sun.

5 0
3 years ago
Read 2 more answers
What is the technology used behind scanning probe microscopes?
Wittaler [7]

hola ya sabes  la respuesta si si ,me la dices porfa

5 0
4 years ago
Read 2 more answers
A 5.0-μC charge is placed at the 0 cm mark of a meter stick and a -4.0 μC charge is placed at the 50 cm mark. At what point on a
Maksim231197 [3]

Answer:

The distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

Explanation:

Given that

q₁ = 5 μ C

q₂ = - 4 μ C

The distance between charges = 50 cm

d= 50 cm

Lets take at distance x from the charge μ C ,the electrical field is zero.

That is why the distance from the charge - 4 μ C =  50 - x cm

We know that ,electric field is given as

E=K\dfrac{q}{r^2}

K\dfrac{5\ \mu}{x^2}=K\dfrac{4\mu }{(50-x)^2}\\\\\dfrac{5}{x^2}=\dfrac{4 }{(50-x)^2}\\\\\\5(50-x)^2=4x^2\\(50-x)^2=0.8x^2\\\\50-x =0.89x\\\ x=\dfrac{50}{1.89}\ cm\\\\\\x=26.45\ cm\\

Therefore the distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

3 0
3 years ago
Other questions:
  • Who is the featured character in an earth diver story
    11·1 answer
  • Do all of them please and explain
    15·2 answers
  • Oil flows upward in the wick of a lantern because of the liquid property called
    6·2 answers
  • For the following situations, describe the contact and non-contact forces that are involved.
    10·2 answers
  • What is the total distance the hiker walked ? A.4 km B.6 km C.10 km D.20 km/
    9·1 answer
  • Which describes the relationship between the frequency wavelength, and speed of a wave as the wave travels through
    9·1 answer
  • What is an electromagnetic wave?
    13·1 answer
  • The Newton is the SI unit for ____.<br> temperature<br> mass<br> weight<br> density
    15·1 answer
  • A bullet is at rest. It travels a distance of 0.34m in a time of 0.0095 seconds. Calculate its acceleration
    12·1 answer
  • A 230 kg steel crate is being pushed along a cement floor. The force of friction is 480 N to the left and the applied force is 1
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!