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Physics Help HERE ONLY !!

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Aodian:

--- Quote from: ManaH on May 26, 2009, 12:02:24 pm ---
--- Quote from: Aodian on May 26, 2009, 11:59:39 am ---Thanx a lot Manah and Shogun!

--- End quote ---
No problem but dont forget + rep  ;)

--- End quote ---
lol no problem.. but still a noob here lol.. so show me how to add rep

Aodian:
nvm i found it lol.. it didnt show earlier but i got it

Bani:

--- Quote from: Shogun on May 25, 2009, 04:20:04 pm ---2a) put different value of masses on bothe sides of the pivot  but make sure tht after using the formula (fxd) (where f=force and d=distance) the product must be same. (then only they are balanced i.e in equilibrium)

 the distance are given for each hole to the pivot. so i recomend to take the first hole(with distance 4cm) and add mass 200g at one side of the pivot. and on the other side,  add mass 100g on the second hole (with disatnce 8 cm from teh pivot)

so when using the formula "fxd " u get on one side: 200x4=800N
                                                     and                                       
                                                   on the other side: 100x8=800N
                         - see both have the same product so it is now they are balanced(in equilibrium)


ans2b) the disc does not move or rotate as u put the mass so they are balanced so no turning effect which means no moment.

c)use formula "fxd"
               200x4=800N
               100x8=800N

d)this is the force of pivot which is downwar to get tht simply add the masses u put
100+800=900g

but rmember u can choose any mass at any distance but the product with equation "fxd" must be same

thts all,hope this helped, any confusion or difficulties or help u need, im all here

--- End quote ---
Hey,the formula for calculating the moment is f*d, and when we multiply 200g by 4cm, aren't we multiplying the mass by the distance? Aren't we first suppossed to convert the mass into force by multiplying it with 10 (F=ma)? and the unit for moment is supposed to be Nm, isn't it? Please help. Thanks in advance :)

hala999:
hey guyz !!!
i need help wid nov 08 1st variant q 5 c

(c) Using a 40 W heater, 16.3 g of ice is melted in 2.0 minutes. The heater is then switched
off. In a further 2.0 minutes, 2.1 g of ice is melted.
Calculate the value of the specific latent heat of fusion of ice from these results.

emi:
L = 40*120/16.3-2.1  = 338.02 J/g

change the minutes to seconds , subtract the first and final mass of ice melted and simple divide with energy

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