Asked by Maggie
On my physics home work i have a question that puts the weight in grams, and i need to find the acceleration. I know that the formula is F=ma, and Force is in newtons, mass is in kg and acceleration is in m/s^2. but can you still find the acceleration by using grams with out converting? because when a converted the mass (1.5g) to kg (.0015kg) and then put it throught the equation the answer didn't fit with the problem.
the problem is asking the acceleration of a bug (1.5g) when a spider pulls it by a massless spider silk with .001N.
when i put in the formula without converting i get .0073m/s^2. which seems plausible.
but when i convert i get 7.3m/s^2. which would make a VERY fast spider....
so i guess my question is just:
do you HAVE to convert in order to get the correct answer
the problem is asking the acceleration of a bug (1.5g) when a spider pulls it by a massless spider silk with .001N.
when i put in the formula without converting i get .0073m/s^2. which seems plausible.
but when i convert i get 7.3m/s^2. which would make a VERY fast spider....
so i guess my question is just:
do you HAVE to convert in order to get the correct answer
Answers
Answered by
Maggie
oops. correction. the spider is pulling with .011N. NOT .001
Answered by
Gene
Maggie, you on the right way but I don't understand how did you get both the values.
The solution is
a = F/m = 1*10^(-3)/1.5*10^(-3)= 1/1.5 = (2/3) m/s^2 = 0.67 m/s^2.
The solution is
a = F/m = 1*10^(-3)/1.5*10^(-3)= 1/1.5 = (2/3) m/s^2 = 0.67 m/s^2.
Answered by
Gene
Not a big deal:
a = 1.1*10^(-2)/1.5*10^(-3) =
= (1.1/1.5)*10 = 7.3 m/s^2
Good for a spider!
a = 1.1*10^(-2)/1.5*10^(-3) =
= (1.1/1.5)*10 = 7.3 m/s^2
Good for a spider!
Answered by
Elena
a= F/m=0.001/1.5•10⁻³=0.67 m/s²
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