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statuscvo
3 months ago
13

Russ makes the diagram below to organize his notes about how Newton’s first law describes objects at equilibrium.

Physics
2 answers:
Ostrovityanka [3.2K]3 months ago
8 0

Correct options:

There is no net force acting on it.

Explanation:

According to Newton's first law, referred to as the law of inertia:

"An object that remains stationary when there is no net force will stay in that state, and an object moving uniformly when there is no net force will continue to move uniformly"

The first scenario (object at stationary) is termed static equilibrium, while the second scenario (object moving uniformly) is termed dynamic equilibrium. As deduced from the description of Newton's first law, both scenarios reflect an object under the influence of a net force of zero, indicating this label belongs in section X (the area shared by both dynamic and static equilibrium).

kicyunya [3.2K]3 months ago
6 0
According to Newton's first law, an object remains at rest until an external force acts upon it, or an object in motion continues to move at a constant speed without accelerating.

Thus, x can solely represent a body coming to a halt. Accordingly, the last option is the most correct.
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A (1.25+A) kg bowling ball is hung on a (2.50+B) m long rope. It is then pulled back until the rope makes an angle of (12.0+C)o
Ostrovityanka [3204]

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

    Em_{f} = K = ½ m v²

    Emo =Em_{f}

    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

   

7 0
3 months ago
Can pockets of vacuum persist in an ideal gas? Assume that a room is filled with air at 20∘C and that somehow a small spherical
kicyunya [3294]

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:

\frac{1}{2}mv^2 = \frac{3}{2}RT

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 = \sqrt{\frac{3RT}{M} }............................1

v = \sqrt{\frac{3(8.314)293}{29*10{-3}kg} }

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 = \frac{r}{v}

which gives us:

time taken = \frac{1*10^{-2}m}{501.99}

so, time taken = 19.92 × 10^{6} seconds = 20μs.

Thus, the required time is 20 μs.

3 0
4 months ago
A major disturbance that caused the ecosystem to stabilize at a new equilibrium?
serg [3582]
The complete removal of all hawks allows for stabilization at a new equilibrium.
7 0
2 months ago
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