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Aliun [14]
2 years ago
9

Migratory birds are able to use the earth's magnetic field to navigate even when clouds and darkness prevent them from having vi

sual references for their flight. The range of sensitivity of such birds extends to magnetic fields as small as about a third of the earth's natural field. If such a bird is flying past a power line which carries 105 amps, and if we assume that the minimum field detectable at 60 Hz is the same as the minimum field detected at DC, at what distance could the bird detect the presence of the power line? Why will the real answer be much smaller? (think about how power lines are arranged)
Physics
1 answer:
stealth61 [152]2 years ago
4 0

Answer:

The receptors that sense the Earth's magnetic field are probably located in the birds' eyes. Now, researchers at Lund University have studied different proteins in the eyes of zebra finches and discovered that one of them differs from the others: only the Cry4 protein maintains a constant level throughout the day and in different lighting conditions

Explanation:

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Assuming the starting height is 0.0 m, calculate the potential energy of the cart after it has been elevated to a height of 0.5
Bogdan [553]
The potential energy is most often referred to as the "energy at rest" and is dependent on the elevation of an object. This can be calculated through the equation,

     E = mgh

where E is the potential energy, m is the mass, g is the acceleration due to gravity, and h is the height. In this item, we are not given with the mass of the cart so we assume it to be m. The force is therefore,

   E = m(9.8 m/s²)(0.5 m) = 4.9m

Hence, the potential energy is equal to 4.9m.
8 0
2 years ago
You are using a hydrogen discharge tube and high quality red and blue light filters as the light source for a Michelson interfer
boyakko [2]

Answer:

final displacement = +24484.5 nm

Explanation:

The path difference when 158 bright spots were observed with red light (λ1 = 656.3 nm) is given as;

Δr = 2d2 - 2d1 = 150λ1

So, 2d2 - 2d1 = 150λ1

Dividing both sides by 2 to get;

d2 - d1 = 75λ1 - - - - eq1

Where;

d1 = distance between the fixed mirror and the beam splitter

d2 = position of moveable mirror from splitter when 158 bright spots are observed

Now, the path difference between the two waves when 114 bright spots were observed is;

Δr = 2d'2 - 2d1 = 114λ1

2d'2 - 2d1 = 114λ1

Divide both sides by 2 to get;

d'2 - d1 = 57λ1

Where;

d'2 is the new position of the movable mirror from the splitter

Now, the displacement of the moveable mirror is (d2 - d'2). To get this, we will subtract eq2 from eq1.

(d2 - d1) - (d'2 - d1) = 75λ1 - 57λ2

d2 - d1 - d'2 + d1 = 75λ1 - 57λ2

d2 - d'2 = 75λ1 - 57λ2

We are given;

(λ1 = 656.3 nm) and λ2 = 434.0 nm.

Thus;

d2 - d'2 = 75(656.3) - 57(434)

d2 - d'2 = +24484.5 nm

5 0
2 years ago
A mover hoists a 50 kg box from the ground to a height of 2 m. What was the change in the box's energy
SSSSS [86.1K]

Answer:

980 J

Explanation:

The change in box's energy is equal to its change in gravitational potential energy:

\Delta U = m g \Delta h

where

m = 50 kg is the mass of the box

g = 9.8 m/s^2 is the acceleration due to gravity

\Delta h= 2m is the change in height of the box

Substituting numbers, we find

\Delta U = (50 kg)(9.8 m/s^2)(2 m)=980 J

3 0
2 years ago
Two resistors of 5.0 and 9.0 ohm are connected in parallel. A 4.0 ohm resistor is then connected in series with the parallel com
rewona [7]

Answer:

I1 = 0.772 A

Explanation:

<u>Given</u>: R1 = 5.0 ohm, R2 = 9.0 ohm, R3 = 4.0 ohm, V = 6.0 Volts

<u>To find</u>:  current I = ? A

<u>Solution: </u>

Ohm's law  V= I R

⇒   I = V / R

In order to find R (total) we first find R (p) fro parallel combination. so

1 / R (p) = 1 / R1  + 1/ R2          ∴(P) stand for parallel

R (p) = R1R2 / ( R1 + R2)

R (p) = (5.0 × 9.0) / (5.0 + 9.0)

R (p) = 3.214 ohm

Now R (total) = R (p) + R3     (as R3 is connected in series)

R (total) = 3.214 ohm + 4.0 Ohm

R (total) = 7.214 ohm

now I (total) = 7.214 ohm / 6.0 Volts

I (total) = 1.202 A

This the total current supplied by 6 volts battery.

as voltage drop across R (p) = V = R (p) × I (total)

V (p) = 3.214 ohm × 1.202 A  = 3.864 volts

Now current through 5 ohms resister  is I1 = V (P) / R1

I1 = 3.864 volts / 5 ohm

I1 = 0.772 A

3 0
2 years ago
A ship maneuvers to within 2.50 x 103 m of anisland's 1.80 x 103 m high mountain peak and fires aprojectile at an enemy ship 6.1
const2013 [10]

Answer:

Distance between peak height (vertically) of projectile and mountain height = (2975.2 - 1800) = 1175.2 m

Distance between where the projectile lands and ship B = (3188.8 - 3110) = 8.8 m

Explanation:

Given the velocity and angle of shot of the projectile, one can calculate the range and maximum height attained by the projectile.

H = (v₀² Sin²θ)/2g

v₀ = initial velocity of projectile = 2.50 × 10² m/s = 250 m/s

θ = 75°, g = 9.8 m/s²

H = 250² (Sin² 75)/(2 × 9.8) = 2975.2 m

Range of projectile

R = v₀² (sin2θ)/g

R = 250² (sin2×75)/9.8

R = 250² (sin 150)/9.8 = 3188.8 m

Height of mountain = 1.80 × 10³ = 1800 m

Maximum height of projectile = 2975.2 m

Distance between peak height (vertically) of projectile and mountain height = 2975.2 - 1800 = 1175.2 m

Distance of ship B from ship A = 2.5 × 10³ + 6.1 × 10² = 2500 + 610 = 3110 m

Range of projectile = 3188.8 m

Distance between where the projectile lands and ship B = 3188.8 - 3110 = 8.8 m

8 0
2 years ago
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