True or False: Molecules in a gas resist crowding and get as far apart as possible. Free electrons also resist crowding and get

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

The string does not experience any force of tension, as it balances two forces acting in the same direction. Hence, the tension is zero.

Explanation:

If tension existed in the string, it would mean that two equal but opposite forces are exerting pull in contrary directions.

When a force of f newtons is applied from the right and another force of f newtons from the left, the resulting action occurs through one force. Because there is action on the same string in opposing directions, the tension in the string can only be equal to the magnitude of the string itself.

Therefore, the string indeed has no tension since it is dealing with two forces acting in the same direction. Thus, the tension is zero.

Answer: 8 m.

Explanation:

Utilize the formula for the perimeter of a rectangle.

Perimeter = 2( l + b)

=2( 3 + 1 )

=8 m

Answer:

The wavelengths of macroscopic objects are too brief for detection.

Explanation:

De Broglie wavelength defines the wavelength of an object as:

where 'h' represents Planck's constant, 'm' denotes the object's mass, and 'v' reflects its velocity.

For macroscopic entities, the mass is significantly higher than that of microscopic ones. Aper the above equation, there's an inverse relationship between an object's mass and its wavelength.

Thus, for significantly large masses, the wavelength is exceedingly short making them untraceable. Consequently, we do not observe wave characteristics in macroscopic objects.