Answer:
220 A
Explanation:
The magnetic force on the floating rod due to the rod held close to the ground is F = BI₁L where B = magnetic field due to rod held close the ground = μ₀I₂/2πd where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, I₂ = current in rod close to ground and d = distance between both rods = 11 mm = 0.011 m. Also, I₁ = current in floating rod and L = length of rod = 1.1 m.
So, F = BI₁L
F = (μ₀I₂/2πd)I₁L
F = μ₀I₁I₂L/2πd
Given that the current in the rods are the same, I₁ = I₂ = I
So,
F = μ₀I²L/2πd
Now, the magnetic force on the floating rod equals its weight , W = mg where m = mass of rod = 0.10kg and g = acceleration due to gravity = 9.8 m/s²
So, F = W
μ₀I²L/2πd = mg
making I subject of the formula, we have
I² = 2πdmg/μ₀L
I = √(2πdmg/μ₀L)
substituting the values of the variables into the equation, we have
I = √(2π × 0.011 m × 0.1 kg × 9.8 m/s²/[4π × 10⁻⁷ H/m × 1.1 m])
I = √(0.01078 kgm²/s²/[2 × 10⁻⁷ H/m × 1.1 m])
I = √(0.01078 kgm²/s²/[2.2 × 10⁻⁷ H])
I = √(0.0049 × 10⁷kgm²/s²H)
I = √(0.049 × 10⁶kgm²/s²H)
I = 0.22 × 10³ A
I = 220 A
the electrons shared between the oxygen and hydrogen atoms spend more time around the oxygen atom nucleus than around the hydrogen atom nucleus
Answer:
<em>1.01 W/m</em>
Explanation:
diameter of the pipe d = 30 mm = 0.03 m
radius of the pipe r = d/2 = 0.015 m
external air temperature Ta = 20 °C
temperature of pipe wall Tw = 150 °C
convection coefficient at outer tube surface h = 11 W/m^2-K
From the above,<em> we assumed that the pipe wall and the oil are in thermal equilibrium</em>.
area of the pipe per unit length A = 
= 
m^2/m
convectional heat loss Q = Ah(Tw - Ta)
Q = 7.069 x 10^-4 x 11 x (150 - 20)
Q = 7.069 x 10^-4 x 11 x 130 = <em>1.01 W/m</em>
Answer:
The specific latent heat of a substance is the amount of energy required to change the state of one kilo of the substance without change in it temperature.The latent heat of vaporization or evaporation is the heat given to some mass to convert if from the liquid to the vapor phase.
10.3
Explanation:
Step 1:
The pressure exerted by any liquid column of height, h density d is given by the formula P = h * d * g
Step 2:
It is given that one atmosphere pressure pushes up 76.0 cm of mercury, we need to calculate the level of water that will be pushed by the same pressure.
Step 3:
Since the pressure pushing up mercury and water is the same
* 
* g = 
* 
* g
= 
= (13.6 g/cm * 76 cm)/1 g/cm = 1033.6 cm
Step 4:
Now we need to express the answer in meters.
1 m = 100 cm.
1033.6 cm = 10.336 m
This can be rounded off to 10.3 m
