Recent content by tuki

So then i have: $$ \Delta S = \int_{T_0}^{T_1} \frac{cm}{T}dT = cm(ln(T_1)-ln(T_0)) $$ now when i compute ##\Delta S## for cold and hot $$ \Delta S_{\text{total}} = cm(ln(T_n)-ln(T_H)+ln(T_c)-ln(T_n)) = cm(ln(T_c)-ln(T_h)) $$ Where ## T_n ## is temperature at balanced state. Now ##\Delta...

## \int \frac{dQ}{T} = \int \frac{mc}{T}dT ## ?

## dQ = mcdT ## ?

Well $$ T = \frac{dQ}{dS} \implies dT = d(\frac{dQ}{dS}) $$ Not quite sure how I would do this? (this one probably not even in right direction)

$$ S = \int \frac{1}{T}dQ $$ but what from now? this equation has ## \Delta T ## but not T $$ Q = cm \Delta T $$ I think I want ## T(Q) ## so I can integrate this properly?

I tried following: $$ dS_{\text{total}} = |\frac{dQ}{T_c}| |\frac{dQ}{T_H}| $$ where ## T_h ## is temperature of hot water and ## T_c ## is temperature of cold water. Coefficient for water wasn't provided in the assignment so i used following value c = 4190 J/kgK. $$ dS_{\text{total}} =...

I have the definition of change in internal energy. $$ \Delta U = Q - W $$ I can get the work by $$ W = \int_{V_1}^{V_2} p dV = p \Delta V $$ however the pressure isn't constant so this won't do. ## W ## is work done by the gas and ## Q ## is amount of heat energy brought into the system. I'm...

Any ideas on how i could formulate this into equation? Could you give a hint? I don't know how i can form equation with final temperature?

Since ΔT is change in temperature, the container and it's contents and the led most have same temperature difference when the led is added. I tried by assuming that energy released by the led is same as the amount that container and it's contents absorb. Meaning Q1-Q2 = 0 => Q1 = Q2. $$ \Delta...

I tried following: $$ \Delta l = \alpha l_0 \Delta T $$ $$ (\Delta l)^2 l_0 = \alpha l_0^2 \Delta T \Delta l $$ $$ \Delta A l_0 = \alpha A_0 \Delta T $$ $$ \Delta A = \frac{ \alpha A_0 \Delta T }{ l_0 } $$ If we remember that: $$ \Delta l = \alpha l_0 \Delta T $$ So we have $$ \Delta A = \frac{...

I tried Bernoulli's equation and ended up with this. $$ p_1 + \frac{1}{2}pv^2_1 = p_2 + \frac{1}{2}pv_2^2 $$ Speed can be expressed as $$ v = \frac{Q}{\pi (\frac{d}{2})^2} $$ $$ \implies v_2 = \sqrt{\frac{p_1-p_2}{\rho}+(\frac{Q}{(\frac{d}{2})}^2 \pi)^2\cdot 2} $$ when i compute with numbers i...

Homework Statement Fire hose has diameter of 4.0 cm and flow rate of 10 L/s. There is pressure of 2.2 bar inside the hose. How high the water can go at best? Water density is 1.00E3 kg/m^3 and air pressure outside the hose is 1.0 bar. Homework Equations Flow rate $$ Q = Av $$ Newtons...

Yes i think it's suppose to be 1.03*10^3 kg/m³. There is mistake in the description of the assignment.

And the sea water density is used is 997kg/m³

The deeper you go the greater the pressure caused by water is. More water above submarine => greater pressure. I don't see how this would affect the buoyancy of the submarine? The force caused by pressure is distributed evenly on top, bottom and sides of the submarine, meaning the sum of the...