a particle has displacement12 cm towards east and 9cm towards north and then 6cm vertically upward. find the magnitude of sum of these displacement
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Answers
Answered by
2
If we draw a rough diagram we get a right angled triangle with sides 12 cm and 15 cm.
displacement is shortest distance,so,here displacement is hypotenuse.
=>S=(12^2+15^2)^1/2
=(144+225)^1/2
=19.2 cm
displacement is shortest distance,so,here displacement is hypotenuse.
=>S=(12^2+15^2)^1/2
=(144+225)^1/2
=19.2 cm
I think this is correct!!
Answered by
0
√261=16.15m
Phytagorilla theorem.
r Horizontal distance travelled will be √(12^2+9^2)=15
cm
Now as vertical displacement is 6 cm,
Total displacement will be √(15^2+6^2)=√261 cm
cm
Now as vertical displacement is 6 cm,
Total displacement will be √(15^2+6^2)=√261 cm
An alternative solution is here first we apply pythagoras thm. on 12 n 9cm so that v
get 15 as resutant... afterward u know that body is 6
cm vertically upward so here we will apply pythagoras
on 15 and so that we get an 3-D image whose
resultant will be √261
get 15 as resutant... afterward u know that body is 6
cm vertically upward so here we will apply pythagoras
on 15 and so that we get an 3-D image whose
resultant will be √261
i hope you will have got it now adesh if need more help then you can ask
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