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Sagot :
a)If it takes the bird 18.0 minutes to fly 6 km away from the earth, the wind's speed will be 4 m/s.
b) The bird would need 7 minutes and 42 seconds to fly back 6 kilometers if he turned around and flew with the wind.
c)Compared to the 133.33 seconds it would take without the wind, the overall round-trip time is affected by the wind.
What is velocity?
The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.
The given data in the problem is
A seagull flies at a velocity,[tex]\rm v_{SA} = 9 \ m/sec[/tex]
The time the bird takes,t=18.0 min
The distance traveled relative to the earth = 6.00 km
a)
The seagull's relative velocity with reference to the ground as;
[tex]\rm v_{sg} = \frac{6.00 \times 10^3 \ m }{(20 min) \times \frac{60 s }{1 \ min}} \\\\ v_{sg}= 5.00 \ m/sec[/tex]
Air velocity with reference to the ground is;
[tex]\rm v_{AG}= v_{SG}-v_{SA} \\\\ v_{AG} = 5.00 \ m/sec - 9.00 \ m/sec \\\\ v_{AG} = -4.00 \ m/sec[/tex]
b)
If the bird turns around and flies with the wind, The time will he take to return 6.00 km is;
[tex]\rm v_{SG}=v_{SA}+v_{AG} \\\\ v_{SG}=-900 \ m/sec +(-4.00 \ m/sec) \\\\ v_{SG}= -13.00 \ m/sec[/tex]
The time the bird takes;
[tex]\rm t = \frac{x_{SG}}{v_{SG}} \\\\ t = \frac{6.00 \times 10^3 \ m }{13.00 \ m/sec } \\\\ t = 462 m/sec \\\\ t = 7 \ min \ and \ 42 \ sec[/tex]
c)\
The total round-trip time compared to what it would be with no wind. is;
[tex]\rm t = 20 \ min( \frac{60 \ sec }{1 \ min} )+ 462 \ sec \\\\ t = 1200 \ sec +6 462 \ ec \\\\ t= 1662 \ sec[/tex]
The time for the round trip is;
[tex]\rm t = \frac{12 \times 10^ 3 }{ 9 \ m/sec } \\\\ t = 1333.33 \ sec[/tex]
Hence the wind's speed, the time bird would need to fly back the total round-trip time will be 4 m/s, 7 minutes and 42 seconds and 1333.33 sec.
To learn more about the velocity, refer to the link: https://brainly.com/question/862972.
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