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 hull type that is best for use on ponds, small lakes and calm rivers is Flat Bottom Hull.
A flat bottomed boat is a boat with a flat bottomed, two-chined hull, which allows it to be used in shallow bodies of water, such as rivers, because it is less likely to ground. The flat hull also makes the boat more stable in calm water.
Answer:
A driver.
Explanation:
Using a driver while at least 350 yds away is better than using a iron, because it will be a waste of the par 4 as it is not as powerful as the driver.
It's gravitational potential energy at the top will roughly equal it's kinetic energy when it was released (a little is lost to air resistance). Note this will assume the release point is zero potential energy. (we are free to define it that way, just letting you know). Gravitational potential energy is mgh.
mgh=25J
h=25J/(0.5kg x 9.81m/s^2) = 5.097m
So it goes about 5.1 meters above the point where it was released