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3 years ago

What planet has the most visible rings

Physics

2 answers:

diamong [38]3 years ago

7 0

Saturn as it has trapped particles in it gravitational pull.

evablogger [386]3 years ago

3 0

Answer:

Saturn

The most prominent and most famous planetary rings in the Solar System are those around Saturn, but the other three giant planets (Jupiter, Uranus, and Neptune) also have ring systems.

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For a constant force the impulse could be increased by

Aleksandr [31]

For constant fore the impulse could be increased by increasing the time for the force.

5 0

3 years ago

Read 2 more answers

What is the frequency of a sound wave commonly called?

WARRIOR [948]

Frequency of a sound wave is commonly referred to as pitch. That is the specialized name for frequency of a sound wave.

Just remember it as pitch.

3 0

3 years ago

At what speed does a falling hailstone travel? Does the speed depend on the distance that the hailstone falls?

Anton [14]

Answer:The speed if hailstone dependly largely on its size. A hailstone with a diameter of 0.39 inches,falls wit a speed of 20mph while a hailstone with 3.1 inches in diameter falls at a speed of 110mph.

No speed does not depend on the distance that the hailstone falls.

Explanation: There are other factors that affect the speed of the falling hailstone apart from its size.They are:

1. Friction between the air and the hailstone

2. Wind condition( windy or moist air)

3. The rate at which it melts falling.

7 0

3 years ago

Ernest Rutherford (the first New Zealander to be awarded the Nobel Prize in Chemistry) demonstrated that nuclei were very small

Vaselesa [24]

Answer:

The final kinetic energy of the Helium nucleus (alpha particle) after been scattered through an angle of 120° is

8.00 x 10-13J

Explanation:

In Rutherford Scattering experiment, the collision of the helium nucleus with the gold nucleus is an ELASTIC COLLISION. This means that the kinetic energy is conserved ( The same before and after the collision).

Thus, the final kinetic energy of the helium nucleus is the same as initial kinetic energy (8.00 x 10^-13Joules)

Although, the kinetic energy is converted to potential energy in Coulomb's law equation.

That is,

1/2(mv^2) = (K* q1q2)/r

Where m is the mass of helium nucleus, v is its colliding velocity, k is electrostatic constant, q1 is the charge on helium nucleus, q2 is the charge on gold nucleus, r is impact parameter

6 0

2 years ago

A coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged con

Tcecarenko [31]

Complete question:

A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).

Required:

What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.

Answer:

The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Explanation:

Given;

charge of the coaxial capable, Q = 8.5 µC = 8.5 x 10⁻⁶ C

length of the conductor, L = 50 m

inner radius, r₁ = 1.304 mm

outer radius, r₂ = 9.249 mm

The magnitude of the electric field halfway between the two cylindrical conductors is given by;

$E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}$

Where;

λ is linear charge density or charge per unit length

r is the distance halfway between the two cylindrical conductors

$r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \ \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm$

The magnitude of the electric field is now given as;

$E = \frac{8.5*10^{-6}}{2\pi(8.85*10^{-12})(5.2765*10^{-3})(50)} \\\\E = 5.793*10^5 \ V/m$

Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

5 0

2 years ago