Answer:
it's depent on height and gravity
Answer:
Average speed will be 48.23 km/h
Explanation:
Let the distance up to hill is = d km
Speed when car goes to hill = 38 km/h
So time required ![t=\frac{distance}{speed}=\frac{d}{38}hour](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Bdistance%7D%7Bspeed%7D%3D%5Cfrac%7Bd%7D%7B38%7Dhour)
Speed when car return from hill = 66 km/h
So time required to return fro hill ![t=\frac{d}{66}h](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Bd%7D%7B66%7Dh)
Total time ![t_{total}=\frac{t}{38}+\frac{t}{66}](https://tex.z-dn.net/?f=t_%7Btotal%7D%3D%5Cfrac%7Bt%7D%7B38%7D%2B%5Cfrac%7Bt%7D%7B66%7D)
Total distance = d+d =2d
So average speed![=\frac{total\ distance}{total\ time}=\frac{2d}{\frac{d}{38}+\frac{d}{66}}=48.23km/h](https://tex.z-dn.net/?f=%3D%5Cfrac%7Btotal%5C%20distance%7D%7Btotal%5C%20time%7D%3D%5Cfrac%7B2d%7D%7B%5Cfrac%7Bd%7D%7B38%7D%2B%5Cfrac%7Bd%7D%7B66%7D%7D%3D48.23km%2Fh)
Answer:
Part a)
h = 0.86 cm
Part b)
Level will increase
Explanation:
Part a)
Mass of the ice cube is 0.200 kg
Now from the buoyancy force formula we know that weight of the ice is counter balanced by buoyancy force on the ice
So here we will have
![mg = \rho V_{displaced} g](https://tex.z-dn.net/?f=mg%20%3D%20%5Crho%20V_%7Bdisplaced%7D%20g)
![V_{displaced} = \frac{m}{\rho}](https://tex.z-dn.net/?f=V_%7Bdisplaced%7D%20%3D%20%5Cfrac%7Bm%7D%7B%5Crho%7D)
![V_{displaced} = \frac{0.200}{1260} = 1.59 \times 10^{-4} m^3](https://tex.z-dn.net/?f=V_%7Bdisplaced%7D%20%3D%20%5Cfrac%7B0.200%7D%7B1260%7D%20%3D%201.59%20%5Ctimes%2010%5E%7B-4%7D%20m%5E3)
now as we know that ice will melt into water
so here volume of water that will convert due to melting of ice is given as
![V\rho_w = m_{ice}](https://tex.z-dn.net/?f=V%5Crho_w%20%3D%20m_%7Bice%7D)
![V = \frac{0.200}{1000} = 2\times 10^{-4} m^3](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B0.200%7D%7B1000%7D%20%3D%202%5Ctimes%2010%5E%7B-4%7D%20m%5E3)
So here extra volume that rise in the level will be given as
![\Dleta V = V - V_{displaced}](https://tex.z-dn.net/?f=%5CDleta%20V%20%3D%20V%20-%20V_%7Bdisplaced%7D)
![\pi r^2 h = 2\times 10^{-4} - 1.59 \times 10^{-4}](https://tex.z-dn.net/?f=%5Cpi%20r%5E2%20h%20%3D%202%5Ctimes%2010%5E%7B-4%7D%20-%201.59%20%5Ctimes%2010%5E%7B-4%7D)
![(\pi (0.039^2) h = 0.41 \times 10^{-4}](https://tex.z-dn.net/?f=%28%5Cpi%20%280.039%5E2%29%20h%20%3D%200.41%20%5Ctimes%2010%5E%7B-4%7D%20)
Part b)
Since volume of water that formed here is more than the volume that is displaced by the ice so we can say that level of liquid in the cylinder will increase due to melting of ice
Calculate the magnetic field strength at the ground. Treat the transmission line as infinitely long. The magnetic field strength is then given by:
B = μ₀I/(2πr)
B = magnetic field strength, μ₀ = magnetic constant, I = current, r = distance from line
Given values:
μ₀ = 4π×10⁻⁷H/m, I = 170A, r = 8.0m
Plug in and solve for B:
B = 4π×10⁻⁷(170)/(2π(8.0))
B = 4.25×10⁻⁶T
The earth's magnetic field strength is 0.50G or 5.0×10⁻⁵T. Calculate the ratio of the line's magnetic field strength to earth's magnetic field strength:
4.25×10⁻⁶/(5.0×10⁻⁵)
= 0.085
= 8.5%
The transmission line's magnetic field strength is 8.5% of that of earth's natural magnetic field. This is no cause for worry.
You just multiply these two numbers. It's 5200J, or 5.2kJ