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Mass of a timber is $20 \ g$. And, density of that timber is $0.27 \ g/cc$. That timber was bind to a metallic materials and, it was released to $0.970 \ g/cc$ water. How much the wood was submer...
#5: Post edited
>Mass of a timber is $20 \ g$. And, density of that timber is $0.27 \ g/cc$. That timber was bind to a metallic materials and, it was released to $0.970 \ g/cc$ water. How much the wood was submerged in water- I was trying to solve the problem following way.
- $$F=Ah\rho g$$
- $$=V\rho g$$
- $$=V \ m^3 \cdot 970 \ kgm^{-3} \cdot 9.8 ms^{-2}$$
- $$=9506V \ N$$
- $$W=mg$$
- $$=20_{\times 10^3} \ kg \times 9.8 \ ms^{-2}$$
- $$=196000 \ N$$
- $$W=F$$
- $$9506V \ N=196000 \ N$$
- $$V=20.6185 \ m^3$$
- I think that I didn't do any mistake while solving that problem. Even, I didn't do any mistake of Units. But, I was wondering why my book wrote that $V=20.62 \ cc$. Both answer matched. But, I was thinking of Units. I am just showing a line what they did than, you will understand what they actually did.
- $$19620 \ dyne = 951.57 V \ dyne$$
- $$V=20.62 \ cm^3=20.62 \ cc$$
- >Mass of a timber is $20 \ g$. And, density of that timber is $0.27 \ g/cc$. That timber was bind to a metallic materials and, it was released to $0.970 \ g/cc$ water. How much the wood was submerged in water?
- I was trying to solve the problem following way.
- $$F=Ah\rho g$$
- $$=V\rho g$$
- $$=V \ m^3 \cdot 970 \ kgm^{-3} \cdot 9.8 ms^{-2}$$
- $$=9506V \ N$$
- $$W=mg$$
- $$=20_{\times 10^3} \ kg \times 9.8 \ ms^{-2}$$
- $$=196000 \ N$$
- $$W=F$$
- $$9506V \ N=196000 \ N$$
- $$V=20.6185 \ m^3$$
- I think that I didn't do any mistake while solving that problem. Even, I didn't do any mistake of Units. But, I was wondering why my book wrote that $V=20.62 \ cc$. Both answer matched. But, I was thinking of Units. I am just showing a line what they did than, you will understand what they actually did.
- $$19620 \ dyne = 951.57 V \ dyne$$
- $$V=20.62 \ cm^3=20.62 \ cc$$
#4: Post edited
>Mass of a timber is $20 \ g$. And, density of that timber is $0.27 \ g/cc$. That timber was bind to a metallic materials and, it was released to $0.970 \ g/cc$ water.- I was trying to solve the problem following way.
- $$F=Ah\rho g$$
- $$=V\rho g$$
- $$=V \ m^3 \cdot 970 \ kgm^{-3} \cdot 9.8 ms^{-2}$$
- $$=9506V \ N$$
- $$W=mg$$
- $$=20_{\times 10^3} \ kg \times 9.8 \ ms^{-2}$$
- $$=196000 \ N$$
- $$W=F$$
- $$9506V \ N=196000 \ N$$
- $$V=20.6185 \ m^3$$
- I think that I didn't do any mistake while solving that problem. Even, I didn't do any mistake of Units. But, I was wondering why my book wrote that $V=20.62 \ cc$. Both answer matched. But, I was thinking of Units. I am just showing a line what they did than, you will understand what they actually did.
- $$19620 \ dyne = 951.57 V \ dyne$$
- $$V=20.62 \ cm^3=20.62 \ cc$$
- >Mass of a timber is $20 \ g$. And, density of that timber is $0.27 \ g/cc$. That timber was bind to a metallic materials and, it was released to $0.970 \ g/cc$ water. How much the wood was submerged in water
- I was trying to solve the problem following way.
- $$F=Ah\rho g$$
- $$=V\rho g$$
- $$=V \ m^3 \cdot 970 \ kgm^{-3} \cdot 9.8 ms^{-2}$$
- $$=9506V \ N$$
- $$W=mg$$
- $$=20_{\times 10^3} \ kg \times 9.8 \ ms^{-2}$$
- $$=196000 \ N$$
- $$W=F$$
- $$9506V \ N=196000 \ N$$
- $$V=20.6185 \ m^3$$
- I think that I didn't do any mistake while solving that problem. Even, I didn't do any mistake of Units. But, I was wondering why my book wrote that $V=20.62 \ cc$. Both answer matched. But, I was thinking of Units. I am just showing a line what they did than, you will understand what they actually did.
- $$19620 \ dyne = 951.57 V \ dyne$$
- $$V=20.62 \ cm^3=20.62 \ cc$$
#3: Post edited
- >Mass of a timber is $20 \ g$. And, density of that timber is $0.27 \ g/cc$. That timber was bind to a metallic materials and, it was released to $0.970 \ g/cc$ water.
- I was trying to solve the problem following way.
- $$F=Ah\rho g$$
- $$=V\rho g$$
$$=V \cdot 970 \cdot 9.8$$- $$=9506V \ N$$
- $$W=mg$$
$$=20_{\times 10^3}\times 9.8$$- $$=196000 \ N$$
- $$W=F$$
$$9506V=196000$$- $$V=20.6185 \ m^3$$
- I think that I didn't do any mistake while solving that problem. Even, I didn't do any mistake of Units. But, I was wondering why my book wrote that $V=20.62 \ cc$. Both answer matched. But, I was thinking of Units. I am just showing a line what they did than, you will understand what they actually did.
- $$19620 \ dyne = 951.57 V \ dyne$$
- $$V=20.62 \ cm^3=20.62 \ cc$$
- >Mass of a timber is $20 \ g$. And, density of that timber is $0.27 \ g/cc$. That timber was bind to a metallic materials and, it was released to $0.970 \ g/cc$ water.
- I was trying to solve the problem following way.
- $$F=Ah\rho g$$
- $$=V\rho g$$
- $$=V \ m^3 \cdot 970 \ kgm^{-3} \cdot 9.8 ms^{-2}$$
- $$=9506V \ N$$
- $$W=mg$$
- $$=20_{\times 10^3} \ kg \times 9.8 \ ms^{-2}$$
- $$=196000 \ N$$
- $$W=F$$
- $$9506V \ N=196000 \ N$$
- $$V=20.6185 \ m^3$$
- I think that I didn't do any mistake while solving that problem. Even, I didn't do any mistake of Units. But, I was wondering why my book wrote that $V=20.62 \ cc$. Both answer matched. But, I was thinking of Units. I am just showing a line what they did than, you will understand what they actually did.
- $$19620 \ dyne = 951.57 V \ dyne$$
- $$V=20.62 \ cm^3=20.62 \ cc$$
#1: Initial revision
Why answer matched but, unit?
>Mass of a timber is $20 \ g$. And, density of that timber is $0.27 \ g/cc$. That timber was bind to a metallic materials and, it was released to $0.970 \ g/cc$ water.
I was trying to solve the problem following way.
$$F=Ah\rho g$$
$$=V\rho g$$
$$=V \cdot 970 \cdot 9.8$$
$$=9506V \ N$$
$$W=mg$$
$$=20_{\times 10^3}\times 9.8$$
$$=196000 \ N$$
$$W=F$$
$$9506V=196000$$
$$V=20.6185 \ m^3$$
I think that I didn't do any mistake while solving that problem. Even, I didn't do any mistake of Units. But, I was wondering why my book wrote that $V=20.62 \ cc$. Both answer matched. But, I was thinking of Units. I am just showing a line what they did than, you will understand what they actually did.
$$19620 \ dyne = 951.57 V \ dyne$$
$$V=20.62 \ cm^3=20.62 \ cc$$