A puck rests on a horizontal frictionless plane. A string is wound around the puck and pulled on with constant force. What fract

Answer:

the time it takes after impact for the puck is 2.18 seconds

Explanation:

initially given information

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thickness = 0.1 mm = 1 × m

dynamic viscosity = 1.75 × Ns/m²

temperature of air = 15°C

to determine

time needed for the puck to reduce its speed by 10%

solution

we note that velocity changes from 0 to v

assuming initial velocity = v

therefore final velocity = 0.9v

implying a change in velocity is du = v

and clearance dy = h

shear stress acting on the surface is expressed as

= µ

therefore

= µ ............1

substituting the values

= 1.75 × ×

= 0.175 v

the area between the air and puck is given by

Area =

area =

area = 7.85 × m²

thus, the force on the puck can be represented as

Force = × area

force = 0.175 v × 7.85 ×

force = 1.374 × v

now applying Newton's second law

force = mass × acceleration

- force =

- 1.374 × v =

solving for t =

the time needed after impact for the puck is 2.18 seconds

1) For x = 6.6 cm,

2) For x = 6.6 cm,

3) For x = 1.45 cm,

4) For x = 1.45 cm,

5) Surface charge density at b = 4 cm:

6) At x = 3.34 cm, the x-component of the electric field equals zero

7) Surface charge density at a = 2.9 cm:

8) None of these regions

Explanation:

1)

The electric field from an infinite charge sheet is perpendicular to it:

where

is the surface charge density

represents vacuum permittivity

Outside the slab, the electric field behaves like that of an infinite sheet.

Consequently, the electric field at x = 6.6 cm (situated to the right of both the slab and sheet) results from the combination of the fields from both:

where

The field from the sheet points left (negative, inward), and the slab’s field points right (positive, outward).

Thus,

and the negative sign indicates a rightward direction.

2)

Both the sheet’s and slab’s fields are perpendicular to their surfaces, directing along the x-axis, hence there's no y-component for the total field.

<pThus, the y-component totals zero.

This happens because both the sheet and slab stretch infinitely along the y-axis. Choosing any x-axis point reveals that the y-component of the field, generated by a surface element dS of either the sheet or slab, , will be equal and opposite to the corresponding component from the opposite side, . Thus, the combined y-direction field is always zero.

3)

This scenario resembles part 1), but the point here is

x = 1.45 cm

which lies between the sheet and the slab. The fields from both contribute leftward as the slab has a negative charge (resulting in an outward field). Thus, the total field computes to

Replacing with expressions from part 1), we get

where the negative illustrates a leftward direction.

4)

This portion parallels part 2). Since both fields remain perpendicular to the slab and sheet, no component exists along the y-axis, thus the electric field's y-component is zero.

5)

Notably, the slab behaves as a conductor, signifying charge mobility within it.

The net charge on the slab is positive, indicating a surplus of positive charge. With the negatively charged sheet on the left of the slab, positive charges shift towards the left slab edge (at a = 2.9 cm), while negative charges move to the right edge (at b = 4 cm).

The surface charge density per unit area of the slab is

This average denotes the surface charge density on both slab sides at points a and b:

(1)

Additionally, the infinite sheet at x = 0 negatively charged , induces an opposite net charge on the slab's left surface, thus

(2)

Having equations (1) and (2) allows for solving the surface charge densities at a and b, yielding:

6)

We aim to compute the x-component of the electric field at

x = 3.34 cm

This point lies inside the slab, bounded at

a = 2.9 cm

b = 4.0 cm

In a conducting slab, the electric field remains at zero owing to charge equilibrium; thus, the x-component thereof in the slab is zero

7)

From part 5), we determined the surface charge density at x = a = 2.9 cm is

8)

As mentioned in part 6), conductors have zero electric fields internally. Since the slab is conductive, the electric field inside remains zero; therefore, the regions where the electric field is null are

2.9 cm < x < 4 cm

Thus, the suitable answer is

"none of these regions"

Learn more about electric fields:

In the study of physics, Hooke's law can be expressed as:

F = kx

This law indicates that the spring force F is proportional to the extension x, with k being the spring constant.

In experiments, this is often examined using the setup illustrated in the included figure. The spring is tested, and a known weight is applied underneath it. This weight exerts a gravitational pull, essentially its weight, on the spring. While the spring elongates, the displacement can be measured using a ruler.

Several potential errors can arise during this experiment. Firstly, the person's measurement reading may be faulty. Digital scales offer greater accuracy as they reduce human error, while ruler readings can be subjective, especially if not viewed at eye level. Additionally, the object's weight may be inaccurately measured if the scale is untrustworthy. Lastly, the measuring equipment may not be correctly calibrated.

Steam transforms into gas as it escapes into the atmosphere. Even if you manage to capture the steam as it ascends, it can revert to H2O when it cools down. Due to the evaporation process, the final volume of water will differ from the original amount.

The heat required to raise the temperature of a substance by is represented by

where m stands for the mass of the substance and indicates the specific heat of the substance. In this situation, we possess and , the specific heat of water.
Consequently, we can ascertain the temperature rise :

Initially, the water's temperature was , so the end temperature should be

Thus, the water is expected to be vapor by now.

However, to give a more accurate statement, during the liquid to vapor transition, the heat added to the system is used to break molecular bonds instead of raising the system's temperature. The heat necessary for the phase change from liquid to vapor is expressed as

where denotes the latent heat of vaporization for water.
Nevertheless, the initial heat input of 50 KJ is less than this requirement, indicating there isn't sufficient heat to finish the liquid-vapor transition. Therefore, the water will remain in the liquid-vapor change phase at a temperature of (the temperature at which the phase change begins)