Answer:
(a)0.0675 J
(b)0.0675 J
(c)0.0675 J
(d)0.0675 J
(e)-0.0675 J
(f)0.459 m
Explanation:
15g = 0.015 kg
(a) Kinetic energy as it leaves the hand

(b) By the law of energy conservation, the work done by gravitational energy as it rises to its peak is the same as the kinetic energy as the ball leave the hand, which is 0.0675 J
(c) The change in potential energy would also be the same as 0.0675J in accordance with conservation law of energy.
(d) The gravitational energy at peak point would also be the same as 0.0675J
(e) In this case as the reference point is reversed, we would have to negate the original potential energy. So the potential energy as the ball leaves hand is -0.0675J
(f) Since at the maximum height the ball has potential energy of 0.0675J. This means:
mgh = 0.0675
0.015*9.81h = 0.0675
h = 0.459 m
The ball would reach a maximum height of 0.459 m
Answer:
a) Not Accurate
b) Not Accurate
c) Accurate
d) Accurate
Explanation:
Part a
Not Accurate, because destructive interference would lead to maximum possible magnitude of < 3 m
Part b
Not Accurate, because constructive interference would lead to minimum possible magnitude of > 2 m
Part c
Accurate, because destructive interference would lead to maximum possible magnitude of < 3 m by varying the phase difference between two waves she can achieve the desired results.
Part d
Accurate, because constructive interference would lead to minimum possible magnitude of > 2 m by varying the phase difference between two waves she can achieve the desired results.
Answer:
If one cup falls down then there will be 59 cups left.
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