Answer:
8.929kg
Explanation:
The work done to place the box at tree house=350 J.
The distance from ground to tree house =4.0 m and ![g=9.8m/s^2](https://tex.z-dn.net/?f=g%3D9.8m%2Fs%5E2)
![E=mgh\\\\m=350J/(9.8m/s^2\times 4.0m)\\\\m=8.929kg](https://tex.z-dn.net/?f=E%3Dmgh%5C%5C%5C%5Cm%3D350J%2F%289.8m%2Fs%5E2%5Ctimes%204.0m%29%5C%5C%5C%5Cm%3D8.929kg)
Hence the mass of the box is 8.929kg
Answer:
![m_i =\frac{1736.702 cal}{129 cal/gr}=13.46 gr](https://tex.z-dn.net/?f=%20m_i%20%3D%5Cfrac%7B1736.702%20cal%7D%7B129%20cal%2Fgr%7D%3D13.46%20gr)
So we need to add 13.46 gr of ice in order to reach the final equilibrium temperature of 49 C
Explanation:
For this case we need to use the fact that the sum for all the heats involved in the system are 0, since we assume an equilibrium state.
Data given
mass of the porcelain cup
the specific heat for the porcelain cup
initial temperature for the coffee and the porcelain cup.
Volume of the coffee.
We can convert this to m^3 and we got 0.000161m^3 and assuming the density fot the coffee equal to the water 1 Kg/m^3 the mass would be:
![m_c = 1 kg/m^3 *0.000161 m^3 = 0.000161 kg=0.161 Kg](https://tex.z-dn.net/?f=%20m_c%20%3D%201%20kg%2Fm%5E3%20%2A0.000161%20m%5E3%20%3D%200.000161%20kg%3D0.161%20Kg)
Specific heat for the coffee
mass of ice required
equilibrium temperature
represent the latent heat of fusin since the ice change the state to liquid.
Solution to the problem
Using this formula:
![\sum_{i=1}^n Q_i = 0](https://tex.z-dn.net/?f=%20%5Csum_%7Bi%3D1%7D%5En%20Q_i%20%3D%200)
We have this:
![m_p cp_p (T_e -T_{ip}) + m_{c} cp_c (T_e -T_{ic}) +m_i L_f + m_i cp_w (T_e -0) =0](https://tex.z-dn.net/?f=%20m_p%20cp_p%20%28T_e%20-T_%7Bip%7D%29%20%2B%20m_%7Bc%7D%20cp_c%20%28T_e%20-T_%7Bic%7D%29%20%2Bm_i%20L_f%20%2B%20m_i%20cp_w%20%28T_e%20-0%29%20%3D0)
Now we can replace and we have this:
![303 gr *(0.260 cal/g C) (49-71)C + 0.161 gr*(1 cal/g C)(49-71)C +m_i [80 cal/gr+(1cal/g C)(49-0)C]=0](https://tex.z-dn.net/?f=%20303%20gr%20%2A%280.260%20cal%2Fg%20C%29%20%2849-71%29C%20%2B%200.161%20gr%2A%281%20cal%2Fg%20C%29%2849-71%29C%20%2Bm_i%20%5B80%20cal%2Fgr%2B%281cal%2Fg%20C%29%2849-0%29C%5D%3D0)
And now we can solve for
and we have:
![-1733.16cal -3.542cal +m_i [129 cal/g]=0](https://tex.z-dn.net/?f=-1733.16cal%20-3.542cal%20%2Bm_i%20%5B129%20cal%2Fg%5D%3D0)
![m_i =\frac{1736.702 cal}{129 cal/gr}=13.46 gr](https://tex.z-dn.net/?f=m_i%20%3D%5Cfrac%7B1736.702%20cal%7D%7B129%20cal%2Fgr%7D%3D13.46%20gr)
So we need to add 13.46 gr of ice in order to reach the final equilibrium temperature of 49 C
A good diet. and exersise is the answer
'D' would do the job ... When you subtract the protons from the mass,
what you have left is neutrons. (The electrons can be ignored. It takes
around 1840 electrons ! to add the mass of a single proton or neutron !)
I don't know it for a fact, but I'd be surprised if the process is really that
simple. I mean, it starts out with knowing the atomic mass, and then
knowing the number of protons in the nucleus. Each of those is a
whole complex problem in itself.
Answer:
A steady decline in driving ability
Explanation:
Alcohol consumption impairs the functioning of the central nervous system. The impairment can range from mood improvement to unconsciousness. As the blood alcohol level (BAL) rises the impairment gets severe. Driving is a complex task that requires the central nervous system to function properly.
Any impairment to the central nervous system leads to a decline in driving activity.