Ofcoure the elephante as the elephant has more weight
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
Answer:
Explanation:
Given
radius of ferris wheel is 20 m
It completes 6 turns in 1 min
i.e. 1 turn in 10 sec
Therefore its angular velocity is 
(a)Period of motion is 10 s
Magnitude of centripetal acceleration is 

(b)Highest point will be 40 m
(c) lowest point 0 m i.e. at ground
Answer:
A. h = 2.15 m
B.
Pb' = 122 KPa
Explanation:
The computation is shown below:
a) Let us assume the depth be h
As we know that

After solving this,
h = 2.15 m
Therefore the depth of the fluid is 2.15 m
b)
Given that
height of the extra fluid is

h' = 0.355 m
Now let us assume the pressure at the bottom is Pb'
so, the equation would be

Pb' = 122 KPa
Answer:
(a) 21.36 ohms
(b) 5.62 A
Explanation:
Parameters given:
Potential difference, V = 120 V
Power, P = 674 W
(a) Power is given as:
P = V²/R
Where R is resistance
=> R = V²/P
R = 120²/674
R = 14400/674
R = 21.36 ohms
(b) Power is also given as:
P = I*V
Where I = Current (time rate of flow of Electric charge)
=> I = P/V
I = 674/120
I = 5.62 A