volume of balloon
= 4/3 T R3
= 4/3 x 3.14 x 6.953
= 1405.47 m3
uplift force
= volume of balloon x density of air x 9.8
= = 1405.47 x 1.29 x 9.8
= 1813.05 x 9.8 N
weight of helium gas
= volume of balloon x density of helium x
9.8
= 1405.47 x .179 x 9.8
= 251.58 x 9.8 N
Weight of other mass = 930 x 9.8 N Total weight acting downwards
= 251.58 x 9.8 +930 x 9.8
= 1181.58 x 9.8 N
If W be extra weight the uplift can balance
1181.58 × 9.8 + W × 9.8 = 1813.05 * 9.8
1181.58+W=1813.05
W= 631.47 kg
The concept required to solve this problem is the optical relationship that exists between the apparent depth and actual or actual depth. This is mathematically expressed under the equations.

Where,
Depth of glass
Refraction index of water
Refraction index of glass
Refraction index of air
Depth of water
I enclose a diagram for a better understanding of the problem, in this way we can determine that the apparent depth in the water of the logo would be subject to



Therefore the distance below the upper surface of the water that appears to be the logo is 4.041cm
Answer:
T_finalmix = 59.5 [°C].
Explanation:
In order to solve this problem, a thermal balance must be performed, where the heat is transferred from water to methanol, at the end the temperature of the water and methanol must be equal once the thermal balance is achieved.

where:

mwater = mass of the water = 0.4 [kg]
Cp_water = specific heat of the water = 4180 [J/kg*°C]
T_waterinitial = initial temperature of the water = 85 [°C]
T_finalmix = final temperature of the mix [°C]

Now replacing:
![0.4*4180*(85-T_{final})=0.4*2450*(T_{final}-16)\\142120-1672*T_{final}=980*T_{final}-15680\\157800=2652*T_{final}\\T_{final}=59.5[C]](https://tex.z-dn.net/?f=0.4%2A4180%2A%2885-T_%7Bfinal%7D%29%3D0.4%2A2450%2A%28T_%7Bfinal%7D-16%29%5C%5C142120-1672%2AT_%7Bfinal%7D%3D980%2AT_%7Bfinal%7D-15680%5C%5C157800%3D2652%2AT_%7Bfinal%7D%5C%5CT_%7Bfinal%7D%3D59.5%5BC%5D)
Answer:
This is your answer
Explanation:
Actually I took this from go ogle