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Aliun [14]
3 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]3 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:

You might be interested in
1.) 450 cm =____m<br><br> 2.) 24 miles =____cm<br><br> 3.) 16.7 gallons =____liters
S_A_V [24]

Answer:

450cm = 4.5m

24miles = 3862430 cm (3.9×10⁶cm)

16.7 gallons = 63.216 liters (approx.)

Explanation:

brainliest plz

7 0
3 years ago
A horizontal force of 40N is needed to pull a 60kg box across the horizontal floor at which coefficient of friction between floo
Mazyrski [523]

Answer:

Coefficient of friction is 0.068.

Work done is 320~J.

Explanation:

Given:

Mass of the box (m): 60 kg

Force needed (F): 40 N

The formula to calculate the coefficient of friction between the floor and the box is given by

F=\mu mg...................(1)

Here, \mu is the coefficient of friction and g is the acceleration due to gravity.

Substitute 40 N for F, 60 kg for m and 9.80 m/s² for g into equation (1) and solve to calculate the value of the coefficient of friction.

40 N=\mu\times60 kg\times9.80 m/s^{2} \\~~~~~\mu=\frac{40 N}{60 kg\times9.80 m/s^{2}}\\~~~~~~~=0.068

The formula to calculate the work done in overcoming the friction is given by

W=Fd..........................(2)

Here, W is the work done and d is the distance travelled.

Substitute  40 N for F and 8 m for d into equation (2) to calculate the work done.

W=40~N\times8~m\\~~~~= 320~J

8 0
3 years ago
You are traveling in a car toward a hill at a speed of 36.4 mph. The car's horn emits sound waves of frequency 231 Hz, which mov
Marina CMI [18]

Answer:

<em>a. The frequency with which the waves strike the hill is 242.61 Hz</em>

<em>b. The frequency of the reflected sound wave is 254.23 Hz</em>

<em>c. The beat frequency produced by the direct and reflected sound is  </em>

<em>    11.62 Hz</em>

Explanation:

Part A

The car is the source of our sound, and the frequency of the sound wave it emits is given as 231 Hz. The speed of sound given can be used to determine the other frequencies, as expressed below;

f_{1} = f[\frac{v_{s} }{v_{s} -v} ] ..............................1

where f_{1} is the frequency of the wave as it strikes the hill;

f is the frequency of the produced by the horn of the car = 231 Hz;

v_{s} is the speed of sound = 340 m/s;

v is the speed of the car = 36.4 mph

Converting the speed of the car from mph to m/s we have ;

hint (1 mile = 1609 m, 1 hr = 3600 secs)

v = 36.4 mph *\frac{1609 m}{1 mile} *\frac{1 hr}{3600 secs}

v = 16.27 m/s

Substituting into equation 1 we have

f_{1} =  231 Hz (\frac{340 m/s}{340 m/s - 16.27 m/s})

f_{1}  = 242.61 Hz.

Therefore, the frequency which the wave strikes the hill is 242.61 Hz.

Part B

At this point, the hill is the stationary point while the driver is the observer moving towards the hill that is stationary. The frequency of the sound waves reflecting the driver can be obtained using equation 2;

f_{2} = f_{1} [\frac{v_{s}+v }{v_{s} } ]

where f_{2} is the frequency of the reflected sound;

f_{1}  is the frequency which the wave strikes the hill = 242.61 Hz;

v_{s} is the speed of sound = 340 m/s;

v is the speed of the car = 16.27 m/s.

Substituting our values into equation 1 we have;

f_{2} = 242.61 Hz [\frac{340 m/s+16.27 m/s }{340 m/s } ]

f_{2}  = 254.23 Hz.

Therefore, the frequency of the reflected sound is 254.23 Hz.

Part C

The beat frequency is the change in frequency between the frequency of the direct sound  and the reflected sound. This can be obtained as follows;

Δf = f_{2} -  f_{1}  

The parameters as specified in Part A and B;

Δf = 254.23 Hz - 242.61 Hz

Δf  = 11.62 Hz

Therefore the beat frequency produced by the direct and reflected sound is 11.62 Hz

3 0
3 years ago
Which type of solar radiation is the most powerful?
cluponka [151]

Answer: Gamma rays

Explanation:

4 0
4 years ago
Read 2 more answers
An object that is initially not rotating has a constant torque of 3.6 N⋅m applied to it. The object has a moment of inertia of 6
Korolek [52]

Answer:

0.6

Explanation:

Angular acceleration is equal to Net Torque divided by rotational inertia, which is the rotational equivalent to Newton’s 2nd Law.  Therefore, angular acceleration is equal to 3.6/6 which is 0.6. Hope this helped!

3 0
2 years ago
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