Density of Object & Oil: Find Out Now!

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The object weighs 285 N in air, 230 N in water, and 261 N in oil. The buoyant force in water is calculated as 55 N, leading to a density of the object of 5.18. The buoyant force in oil is 24 N, resulting in an oil density calculation of 0.436 times that of water. Participants confirm the answers but emphasize the importance of including units in the calculations. Clarity on unit requirements may resolve any confusion regarding the results.
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Homework Statement


An object weighs 285 N in the air. When tied to a string, connected to a balance, and immersed in water, it weighs 230 N. When it is immersed in oil, it weighs 261 N.

(a) Find the density of the object.

(b) Find the density of the oil.

Homework Equations



A=mgd(water)

The Attempt at a Solution




this weight is equal to 285-230=55N corresponding to a density =1
so the density of the body is 285/55=5.18


the weight of oil is 285-261= 24 =Mgd(oil)
compared to water d(oil)/d(water)=24/55=0.436
 
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Your answers are correct, but you should write down the units.
 
I believe my answer is wrong. If someone would double check it.
 
Leo34005 said:
I believe my answer is wrong. If someone would double check it.

I did double check it when I first posted. Your answers are right but missing units. If your teacher wanted different units then perhaps therein lies your confusion.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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