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:
g = 11.2 m/s²
Explanation:
First, we will calculate the time period of the pendulum:
where,
T = Time period = ?
t = time taken = 135 s
n = no. of swings in given time = 98
Therefore,
T = 1.38 s
Now, we utilize the second formula for the time period of the simple pendulum, given as follows:
where,
l = length of pendulum = 54 cm = 0.54 m
g = acceleration due to gravity on the planet = ?
Therefore,
<u>g = 11.2 m/s²</u>
Answer:
Potential energy
Explanation:
Before release, the catapult has potential energy stored in a tension of torsion device in it. Normally a flexible bow like object that could be made of wood or of metal.
Answer:
the motion that repeat itself in equal interval of time is called periodic motion and it is equal to harmonic motion. for example pendulum
Answer:
θ=108rad
t =10.29seconds
α=-8.17rad/s²
Explanation:
Given that
At t=0, Wo=24rad/sec
Constant angular acceleration =30rad/s²
At t=2, θ=432rad as it try to stop because the circuit break
Angular motion
W=Wo+αt
θ=Wot+1/2αt²
W²=Wo²+2αθ
We need to find θ between 0sec to 2sec when the wheel stop
a. θ=Wot+1/2αt²
θ=24×2+1/2×30×2²
θ=48+60
θ=108rad.
b. W=Wo+αt
W=24+30×2
W=84rad/s
This is the final angular velocity which is the initial angular velocity when the wheel starts to decelerate.
Wo=84rad/sec
W=0rad/s, because the wheel stop at θ=432rad
Using W²=Wo²+2αθ
0²=84²+2×α×432
-84²=864α
α=-8.17rad/s²
It is negative because it is decelerating
Now, time taken for the wheel to stop
W=Wo+αt
0=84-8.17t
-84=-8.17t
Then t =10.29seconds.
a. θ=108rad
b. t =10.29seconds
c. α=-8.17rad/s²
