A 65kg person throw a 0.045kg snowball forward with a ground speed of 30m/s. A second person, with a mass of 60kg, catches the s
1 answer:
Well, st first we should find <span>initial momentum for the first person represented in the task which definitely must be :
</span>
And then we find the final one :
Then equate them together :
So we can get the velocity, which is
In that way, according to the main rules of <span>conservation of momentum you can easily find the solution for the second person.
Regards!</span>
</span>
And then we find the final one :
Then equate them together :
So we can get the velocity, which is
In that way, according to the main rules of <span>conservation of momentum you can easily find the solution for the second person.
Regards!</span>
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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.
<h3>What is velocity?</h3>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,
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;
Air velocity with reference to the ground is;
b)
If the bird turns around and flies with the wind, The time will he take to return 6.00 km is;
The time the bird takes;
c)\
The total round-trip time compared to what it would be with no wind. is;
The time for the round trip is;
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: brainly.com/question/862972.
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Answer:
Explanation:
q = Charge
r = Distance
The electric field is given by
The electric field at the aircraft is