@@PhysicsNinja my bad....the question which I've got is a dog of mass 10kg is standing on a flat 10m long boat so that it is 20m from the shore. It walks 8m on the boat towards the shore and then stops. The mass of the boat is 40kg and friction between the boat and the water is negligible. how far is the dog from the shore now?......the actual answer is 13.6m but by using your method i got 14m ......actually you didn't calculate the backward motion of the boat that makes the position of center of mass of boat constant i suppose....
literally no way to visualize wtf you mean by dog relative to ground minus boat relative to ground. that doesnt do anythingto explain even with the picture it doesnt tie anyhing together for the viewer.
listen, lemme tell you something, all these physics channel morons don't know anything about physics, they're just book smart, they're not capable of thinking outside the box, they've memorized every single problem and how to do it whether the method is right or wrong. if this was actual real life, and this exact situation played out, his answer would be completely incorrect.
The answer is not correct...bz dog & boat are moving opposite to each other from groud observation... So the eqn would be:- 2.4= ◇x(d,G) - { - ◇x(B,G) } So 2.4= ◇x(d,G) + ◇x(B,G)
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Thank you so much for this, you showed me several key ideas I was missing from this unit. Excellent explanation
I’ve changed the way I solve these problem. This review has 2 similar problems Center of Mass Review Problems
ruclips.net/video/LFQwhIIAMKw/видео.html
This is my exact problem it's like you got it from the textbook we are using thanks for the help
Nice Explanation Sir. Many Many Sincere Thanks.
Thanks
Classical physics has the lamest problems.
he complicated it fr but he helped. Its just his way of solving has many shortcuts
shouldn't mdxd be negative since it is moving in the left direction?
I had this exact physics problem in my hw
THANK YOU SOO MUCH , HELPED ME A LOT ,
Ah, because the boat moves? I had originally thought 3.7 assuming the boat didn't move.
something is not correct....check it again
what do you think is wrong with the solution? explain and i can help
@@PhysicsNinja my bad....the question which I've got is a dog of mass 10kg is standing on a flat 10m long boat so that it is 20m from the shore. It walks 8m on the boat towards the shore and then stops. The mass of the boat is 40kg and friction between the boat and the water is negligible. how far is the dog from the shore now?......the actual answer is 13.6m but by using your method i got 14m ......actually you didn't calculate the backward motion of the boat that makes the position of center of mass of boat constant i suppose....
@@PhysicsNinja you never explained 3 years late on an explanation I need tonight because of s test tomorrow
Thank you!
made my day...
PREPARING FOR IIT ??
@@humanhuman9228 i am
Thanks❤️❤️❤️❤️❤️
literally no way to visualize wtf you mean by dog relative to ground minus boat relative to ground. that doesnt do anythingto explain even with the picture it doesnt tie anyhing together for the viewer.
listen, lemme tell you something, all these physics channel morons don't know anything about physics, they're just book smart, they're not capable of thinking outside the box, they've memorized every single problem and how to do it whether the method is right or wrong. if this was actual real life, and this exact situation played out, his answer would be completely incorrect.
The answer is not correct...bz dog & boat are moving opposite to each other from groud observation...
So the eqn would be:-
2.4= ◇x(d,G) - { - ◇x(B,G) }
So 2.4= ◇x(d,G) + ◇x(B,G)
4.18 or 4.2 final answer
this is so lame, what ever happened to intuition?