Ask your own question, for FREE!
Physics 8 Online
OpenStudy (anonymous):

Hii can someone help me?

OpenStudy (anonymous):

OpenStudy (anonymous):

hei.., a = -3.2953 m/s^2.., right??

OpenStudy (anonymous):

noo acceleration is -2455 :/

OpenStudy (anonymous):

thats why i said it would be easier for you if you looked at the other link! sorry :/

OpenStudy (anonymous):

okay...., i try again...,

OpenStudy (anonymous):

im sorry :/

OpenStudy (anonymous):

sorry for what??? what is the real answer, on ur book??

OpenStudy (anonymous):

for bothering you! and the real answer for the whole question? .048 m

OpenStudy (anonymous):

@natasha.aries ...,

OpenStudy (anonymous):

yess

OpenStudy (anonymous):

my answer for a is -2399...,

OpenStudy (anonymous):

how?

OpenStudy (anonymous):

@gerryliyana

OpenStudy (mayankdevnani):

the speed is related to energy .5*m*v^2=m*g*h the acceleration of the pad is 0=v-a*t solve for a and plug into s=v0*t-.5*a*t^2 Solution: v=sqrt(2*9.81*12) a=sqrt(2*9.81*12)*1000/6.25 a=2455 m/s^2 s=sqrt(2*9.81*12)*6.25/1000- .5*2455*6.25^2/1000^2 4.8 cm

OpenStudy (anonymous):

ok first of all for g =10 v = sqrt(2gh) = sqrt(2x10x12) = sqrt(240) = 15.49 m/s based on Law of conservatio of energy.., W = 1/2 m (v2^2 - v1^2) m a s = 0.5 m (0 - v^2) a s = -0.5 (15.49^2) a = - 0.5 x239.9401/0.05 = -2399

OpenStudy (mayankdevnani):

ok @natasha.aries

OpenStudy (mayankdevnani):

am i right @jiteshmeghwal9

OpenStudy (anonymous):

yes ur eight @mayankdevnani :)

OpenStudy (mayankdevnani):

thnx...... @gerryliyana

OpenStudy (jiteshmeghwal9):

yes @mayankdevnani :)

OpenStudy (mayankdevnani):

thnx....

OpenStudy (anonymous):

but the answer is .048 cm?

OpenStudy (mayankdevnani):

i don't think

OpenStudy (anonymous):

maybe we just need to change it all into grams!

OpenStudy (mayankdevnani):

may be

OpenStudy (anonymous):

THANK YOU SO MUCH :)

OpenStudy (mayankdevnani):

welcome

OpenStudy (anonymous):

BOTH OF YALL DESERVE THE MEDAL!

OpenStudy (mayankdevnani):

oh!!!

OpenStudy (anonymous):

@natasha.aries didn't we solve this one the other day?!

OpenStudy (anonymous):

yeah but when i was trying to do one part i kept messing up so i wanted to know what i was doing wrong!

OpenStudy (anonymous):

oh ok. Did you notice that mayankdevnani did it slightly differently at the end? There is sometimes more than more way of getting an answer, as long as you use correct physics! hope you are still feeling like yuo are making progress

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!