g Suppose Howard is pulling a bucket of bricks up along the side of a building with a rope. The bricks have a mass of 20 kg and

We will use the equations of rotational kinematics,

            (A)

                                    (B)                                          

Here, and denote the final and initial angular displacements, respectively, whereas and represent final and initial angular velocities, and is the angular acceleration.

We are provided with and .

By substituting these values into equation (A), we have

Now, using equation (B),

This indicates that the wheel's angular speed at the 4.20-second mark is 36.7 rad/s.

Response:

1 slit width = 0.158 mm, slit separation = 0.633 mm, distance between diffraction maxima = 12.7 mm

Explanation:

This involves defining several terms:

λ, the wavelength of the beam

D, the distance from the plane to the slit

x, the distance between minima in the diffraction pattern (in single slit setups)

w, the fringe width for double slit setups

1. In a double slit experiment, the fringe width is also recognized as the distance between Maxima.

Thus, w=λD/d, leading to d=λD/w when rearranged.

So, d= (632.8 x 10^-9 x 1 x 10^3)/1

which equals d= 0.633mm.

For single slit diffraction, minima are defined by a*sinΘ= m*x

where a is the slit width and m is an integer.

For small angles, it simplifies to Θ= (x/D) = (λ/a), allowing us to solve for a:

a = λD/a

a = (632.8x10^-9 x 1)/4

yielding a= 0.158mm.

2. A grating with 20 slits/mm gives d= 1/20mm= 0.05x10^-3.

To find y (the distance between maxima), apply y= λL/d:

Finally, y= (632.8 x 10^-9 x 1)/0.05x10^-3, which results in y= 12.7mm.

The answer is 195 J.

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

Part a) The package's speed matches the helicopter's speed in the horizontal direction. Thus, after a time "t", the horizontal velocity remains constant, while in the Y-direction, it begins to fall under gravity. Part b) The distance relative to the helicopter is equivalent to the distance it falls freely. Part c) If the helicopter is ascending uniformly, the package's final speed after time t can be described in terms of its initial speed and gravity.