Physics, 13.07.2019 22:00 destinywiggins75

Aseagull flies at a velocity of 9.00 m/s straight into the wind. (a) if it takes the bird 20.0 min to travel 6.00 km relative to the earth, what is the velocity of the wind? (b) if the bird turns around and flies with the wind, how long will he take to return 6.00 km? (c) discuss how the wind affects the total round-trip time compared to what it would be with no wind.

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Answer from: emocow

The entire time period of the journey will be 22.222 min without the wind, while it will be (20+7.692) 27.692 mins with the wind flowing.

What is relative velocity?

Relative velocity is the velocity of an object with respect to another.

If it takes the bird 20.0 min to travel 6.00 km relative to the earth, what is the velocity of the wind?

Distance traveled by the bird = 6 km

Time is taken by the bird = 20 min

We know that wind is going in the opposite direction of the bird, therefore, it will provide a drag to the bird,

$(v_{bird} - v_{wind}) = \dfrac{distance}{Time}\\$

$(9 - v_{wind}) = \dfrac{6 \times 1000}{20 \times 60}\\\\v_{wind} = 4\rm\ m/s$

Hence, the speed of the wind is 4 m/s.

If the bird turns around and flies with the wind, how long will it take to return 6.00 km?

Distance traveled by the bird = 6 km

We know that the wind and the bird will be going in the same, therefore, it will increase the bird's speed,

$(v_{bird} + v_{wind}) = \dfrac{distance}{Time}\\$

$(9+4) = \dfrac{6 \times 1000}{Time}\\$

${Time}= 461.538\rm\ sec = 7.692\ min$

Discuss how the wind affects the total round-trip time compared to what it would be with no wind.

Total distance = 12 km

Since there is no wind the speed during the entire journey will be the same,

$v_{bird} = \dfrac{distance}{time}$

$9= \dfrac{12 \times 1000}{time}$

time, t = 1333.3334 sec = 22.222 min

Hence, the entire time period of the journey will be 22.222 min without the wind, while it will be (20+7.692) 27.692 mins with the wind flowing.

Learn more about Relative Velocity:

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Answer from: kmacho9726

Here we will the speed of seagull which is v = 9 m/s

this is the speed of seagull when there is no effect of wind on it

now in part a)

if effect of wind is in opposite direction then it travels 6 km in 20 min

so the average speed is given by the ratio of total distance and total time

$v_{avg} = \frac{6000}{20*60}$

$v_{avg} = 5m/s$

now since effect of wind is in opposite direction then we can say

$V_{net} = v_{bird} - v_{wind}$

$5 = 9 - v_{wind}$

$v_{wind}= 4 m/s$

Part b)

now if bird travels in the same direction of wind then we will have

$v_{net}= v_{bird} + v_{wind}$

$v_{net} = 9 + 4 = 13 m/s$

now we can find the time to go back

$time = \frac{distance}{speed}$

$time = \frac{6000}{13}$

time = 7.7 minutes

Part c)

Total time of round trip when wind is present

T = t_1 + t_2

T = 20 + 7.7 = 27.7 min

now when there is no wind total time is given by

$T = \frac{6000}{9} + \frac{6000}{9}$

T = 22.22 min

So due to wind time will be more

Answer from: Quest

19 hertz bruh u trippin? ?

Answer from: Quest

Amagnetic field "does not" interact with stationary electric charges. so, a.hope my answer you : )

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