Look at the diagram of the moon and three different paths that it could take. A planet with a moon in orbit. The moon has arrows

The duration required for the seventh car to pass amounts to 13.2 seconds. The train's movement is characterized by uniform acceleration, enabling the application of suvat equations. Initially, we analyze the movement of the first car, utilizing the equation for distance s covered in time t, which corresponds to the length of one car, with u = 0 as the initial velocity and a representing acceleration, over t = 5.0 s. We can rearrange the equation reflecting L as the length of one car. This is similarly applicable for the initial seven cars, accounting for the distance of 7L and the required time t'. With constant acceleration retained, we can derive t' through substitution in the equation, leading to fundamental conclusions regarding the relationship exhibited in the graph of distance against time in uniformly accelerated motion.

The formula to apply is expressed as:

ρ = nA/VcNₐ
where
ρ signifies density
n represents the number of atoms within a unit cell (for FCC, n=4)
A indicates the atomic weight
Vc stands for the cubic cell volume equal to a³, with a being the side length (for FCC, a = 4r/√2, where r is the radius)\
Nₐ denotes Avogadro's number, which is 6.022×10²³ atoms/mol

Calculating the radius: r = 0.1387 nm*(10⁻⁹ m/nm)*(100 cm/1m) = 1.387×10⁻⁸ cm
Then find a: a = 4(1.387×10⁻⁸ cm)/√2 = 3.923×10⁻⁸ cm
Then calculate V: V = a³ = (3.923×10⁻⁸ cm)³ = 6.0376×10⁻²³ cm³

Now compute density:
ρ = [(4 atoms)(195.08 g/mol)]/[(6.0376×10⁻²³ cm³)(6.022×10²³ atoms/mol)]Finally, we get ρ = 21.46 g/cm³

Answer:

176.38 rpm

Explanation:

The proportion of mass for arms and legs is 13%.

For legs and trunk, it's 80% of the total mass.

Additionally, the head accounts for 7% of the total mass.

Overall, the mass of the skater is 74.0 kg.

Each arm length is 70 cm, or 0.7 m.

The skater's height is 1.8 m, and the trunk diameter measures 35 cm, or 0.35 m.

Starting angular momentum is 68 rpm.

We make the following assumptions:

1. The skater is modeled as a vertical cylinder (head, trunk, and legs), with arms extending horizontally as two uniform rods.
2. Friction between the skater and the ice is considered negligible.

To analyze her body, we divide it into two parts, treating the arms as spinning rods.

1. Each arm (rod) has a moment of inertia of .

The arms comprise 13% of 74 kg, calculated as 0.13 x 74 = 9.62 kg.

Each arm is then evaluated as 9.62/2 = 4.81 kg.

Let L represent the length of each arm.

Thus,

I = x 4.81 x = 0.79 kg-m for each arm.

2. The body, treated as a cylinder, has a moment of inertia of .

For the body, the radius r is half of the trunk diameter: r = 0.35/2 = 0.175 m.

The mass of the trunk amounts to (80% + 7%) of 74 kg, which calculates to 0.87 x 74 = 64.38 kg.

Therefore, I = x 64.38 x = 0.99 kg-m.

Two cases are considered:

case 1: Body spinning with arms extended.

Total moment of inertia equals the combined moments of inertia of both arms and the trunk.

I = (0.79 x 2) + 0.99 = 2.57 kg-m.

The angular momentum is given by Iω.

Here, ω = angular speed = 68.0 rpm = x 68 = 7.12 rad/s.

The angular momentum then becomes 2.57 x 7.12 = 18.29 kg-rad/m-s.

case 2: Arms drawn in alongside the trunk.

The moment of inertia is attributed solely to the trunk. This is 0.91 kg-m.

The angular momentum equals Iω.

= 0.99 x ω = 0.91ω.

By the principle of conservation of angular momentum, the two angular momentum quantities are equal. Therefore,

18.29 = 0.99ω.

Solving gives ω = 18.29/0.99 = 18.47 rad/s.

This leads to 18.47 ÷ = 176.38 rpm.

We have a mercury atom in its ground state that absorbs 20 eV of energy, resulting in ionization by the loss of an electron. We need to compute the kinetic energy of the ejected electron. The initial energy here is 20 eV, which converts to 20 J/C. The charge of the electron is 1.60217662 × 10^{-19} coulombs. To find the kinetic energy, we can utilize the equation: KE = 20 Joules / Coulombs * 1.60217662 × 10^{-19} coulombs, yielding KE = 1.25x10^{20} Joules. Thus, after ionization, the electron possesses a kinetic energy of 1.25x10^{20} Joules or 1.25x10^{17} kJ.

Response:

The magnetic flux across the surface amounts to Wb

Clarification:

Provided:

Strength of magnetic field T

Circle's radius m

Angle of incidence between the field and the surface normal 25°

Utilizing the flux formula,

Where the angle represents the orientation of the magnetic field line relative to the surface normal, and the area pertains to the circular surface.

The calculation for magnetic flux is expressed as

Wb

Thus, the magnetic flux through the surface calculates to Wb