Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of -2.0 µC; sphere B carries a charge of -6.0 µC;

1. Independent variable: the variable that can be modified and regulated.
the nail polish on Sarah's nails
2. Dependent variable: outcomes that result from the changes in the independent variable.
the duration of the nail polish's longevity
3. <span> Hypothesis: Different brands of nail polish have varied durations before they chip.
</span> 4. Control group: the <span> independent variable remains unchanged in this setup, not subject to variations.
</span> the schedule of when Sarah applies her nail polish (Sarah colors her nails every Sunday for a month)
the specific base coat and top coat (she <span> applies the same bottom coat and top coat with every kind of nail polish)
weekly habits (she ensures the same routine each week so her nails are not treated more harshly on some weeks).
</span> Experimental group: <span> the independent variable is altered for this group
type of nail polish (Essie, OPI, and Sally Hansen)
</span> 6. Constants: the experimenter (Sally), duration of study (one week), her weekly routine, <span> base coat and top coat, </span>

Answer:

The time required is 20 μs

Explanation:

Here is the data provided:

temperature = 20°C  = 293 K

radius = 1 cm

atomic mass of air = 29 u

To determine

the duration for air to refill the vacuum space

solution:

We calculate the root mean square velocity of air particles. This can be expressed as:

where m indicates mass, t is temperature, v is speed, and R is the ideal gas constant, which is approximately 8.3145 (kg·m²/s²)/K·mol.

v = ............................1

v =

Resulting in v = 501.99 m/s.

<pNow, to cover the distance of 1 cm,<pThe duration needed for air is calculated as:

time taken =

which gives us:

time taken =

so, time taken = 19.92 × seconds = 20μs.

Thus, the required time is 20 μs.

Answer:

F = 0.535 N

Explanation:

We will apply energy concepts, considering both the peak and the bottom of the path.

Top

   Em₀ = U = mg y

Bottom

    = K = ½ m v²

    Emo =

    mg y = ½ m v²

    v = √ (2gy)

   y = L - L cos θ

  v = √ (2g L (1 - cos θ))

Next, we will employ Newton's second law at the lowest point where the acceleration is centripetal.

     F = ma

     a = v² / r

For the turning radius, the cable length is r = L.

    F = m 2g (1 - cos θ)

Now, let's find the result.

    F = 2  1.25  9.8 (1 - cos 12)

    F = 0.535 N