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kodGreya [7K]
3 years ago
14

an object has a momentum of 250 kg m/s what would be the momentum of an object that has half of the mass and going at the same v

elovcity
Physics
1 answer:
Romashka-Z-Leto [24]3 years ago
3 0

The momentum of the second object is 125 kg m/s

Explanation:

The momentum of an object is given by

p=mv

where

m is the mass of the object

v is its velocity

For the object in this problem,

p = 250 kg m/s

And its mass is m and its velocity is v.

The second object has a mass of m' = \frac{m}{2} and same velocity v, so its momentum is

p'=m'v = (\frac{m}{2})v=\frac{1}{2}(mv)=\frac{1}{2}p

So, the second object has a momentum that is half of the momentum of the first object, therefore it is:

p'=\frac{1}{2}(250)=125 kg m/s

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

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Scorpion4ik [409]

Part a:

Q_{1} = 56

Q_{2} = 60

Q_{3} = 63

     The quartiles are found by finding the medium of the data, and then the mediums of the two different data sets on either side of the medium. The Q_{2} is the overall medium, Q_{1} is the medium of the first half, and Q_{3} is the medium of the second half.

-> How is the medium found? When finding the medium we put the values in order least to greatest and pick the middle value.

[] See attached

Part b:

The range is 7.

The interquartile range is the range of numbers between Q_{1} and Q_{3}. In other words, it is 50% of the data, directly in the middle.

This becomes 63 - 56 = 7

Part c:

79 is an outlier.

It is an outlier because it is 1.5 above or below (in this case, above) the interquartile range.

-> 63 + (7 + \frac{7}{2}) ≤ 79

-> 63 + 10.5 ≤ 79

-> 73.5 ≤ 79

Have a nice day!

     I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly.

- Heather

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