Answer:
A negatively charged particle -q is placed at the center of a uniformly charged ring, where the ring has a total positive charge Q as shown in the following figure. The particle, confined to move along the x axis, is moved a small distance x along the axis ( where x << a) and released. Show that the particle oscillates in simple harmonic motion with a frequency given by,
Answer:
The answer is "Including all three studies of 0s to 2s, that shift in momentum is equal".
Explanation:
Its shift in momentum doesn't really depend on the magnitude of its cars since the forces or time are similar throughout all vehicles.
Let's look at the speed of the car

We use movies and find lips

The moment is defined by

The moment change

Let's replace the speeds in this equation

They see that shift is not directly proportional to the mass of cars since the force and time were the same across all cars.
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Answer:
I = 0.002593 A = 2.593 mA
Explanation:
Current density = J = (3.00 × 10⁸)r² = Br²
B = (3.00 × 10⁸) (for ease of calculations)
The current through outer section is given by
I = ∫ J dA
The elemental Area for the wire,
dA = 2πr dr
I = ∫ Br² (2πr dr)
I = ∫ 2Bπ r³ dr
I = 2Bπ ∫ r³ dr
I = 2Bπ [r⁴/4] (evaluating this integral from r = 0.900R to r = R]
I = (Bπ/2) [R⁴ - (0.9R)⁴]
I = (Bπ/2) [R⁴ - 0.6561R⁴]
I = (Bπ/2) (0.3439R⁴)
I = (Bπ) (0.17195R⁴)
Recall B = (3.00 × 10⁸)
R = 2.00 mm = 0.002 m
I = (3.00 × 10⁸ × π) [0.17195 × (0.002⁴)]
I = 0.0025929449 A = 0.002593 A = 2.593 mA
Hope this Helps!!!
The density of an object is given by:
D = M/V
D = density, M = mass, V = volume
The volume of the cube V is given by:
V = s³
where s = side length
Make a substitution:
D = M/s³
Given values:
M = 20kg, s = 20×10⁻²m
Plug in and solve for D:
D = 20/(20×10⁻²)³
D = 2500kg/m³