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:
the change in momentum = Force x change in time
Answer:
a = 2 [m/s^2]
Explanation:
To solve this problem we must use the expressions of kinematics, we must bear in mind that when a body is at rest its velocity is zero.
where:
Vf = final velocity = 0
Vi = initial velocity = 60 [m/s]
a = desacceleration [m/s^2]
t = time = 30 [s]
Note: the negative sign of the above equation means that the car is slowing down, i.e. its speed decreases.
0 = 60 - (a*30)
a = 2 [m/s^2]
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Answer:
Zeros to the left of a decimal can be insignificant place holders, such as in 0.043 (two significant figures).
They can be significant if they are between two digits who themselves are significant, such as in 101.000 (three significant figures).
In the case of a number like 1,000 we can see there is only one significant figure. The zero digits are not between sigfigs.
