The Moon orbits Earth in a nearly circular orbit (mean distance is 378,000 km ). The moon Charon orbits Pluto in a nearly circul

Thanks for asking your question here. I hope this response provides clarity. Feel free to ask additional questions. The moment resulting from the two forces about point O is 376 lb-ft counterclockwise.

Answer:

Explanation:

The distance between the electrodes is denoted as d.

The kinetic energy of the electron is represented as Ek when the electrodes are positioned at a distance of "d" apart.

Our goal is to determine the kinetic energy when they are separated by a distance of d/3.

K.E = ½mv²

It’s important to note that the mass remains constant; only velocity varies.

Additionally,

K.E = Work done by the electron

K.E = F × d

K.E = W = ma × d

Assuming constant acceleration

Hence, m and a are fixed,

therefore,

K.E is directly related to d

Thus, as d increases, K.E increases, and conversely, when d decreases, K.E decreases.

Consequently,

K.E_1 / d_1 = K.E_2 / d_2

With K.E_1 equating to E_k

and d_1 being d

while d_2 is represented as d/3

This leads to K.E_2 = K.E_1 / d_1 × d_2

Thus, K.E_2 = E_k × ⅓d / d

Finally,

K.E_2 = ⅓E_k

Therefore, the resultant kinetic energy is one third of the original E_k

Referencing the diagram below, we can deduce from the geometry that x = 2.5 - 0.55 = 1.95 m, leading to cos θ = 1.95/2.5 = 0.78. Therefore, θ = cos⁻¹ 0.78 = 38.74°. According to the free body diagram, the tension in the chain measures 450 N. Here, F denotes the centripetal force and W signifies Dee's weight. The tension's components are as follows: Horizontal component = 450 sin(38.74°) = 281.6 N, directed to the left, and Vertical component = 450 cos(38.74°) = 351.0 N, directed upward. Answers: Horizontal: 281.6, directed left. Vertical: 351.0 N, directed upward.

Response:

The ball remained airborne for 3.896 seconds

Explanation:

Given that

g = 9.8 m/s², representing gravitational acceleration,

If the angle of launch is 45°, the horizontal range will be maximized.

Both horizontal and vertical launch velocities are equal, each equating to

v_h  =  v cos θ

v_h  =  27 × cos 45°

         = 19.09 m/s.

The duration to reach maximum height is half of the flight time.

v = u + at   ∵ v = 0 (at maximum height)

19.09 - 9.8 t₁ = 0

t₁ = 1.948 s

The total time in the air equals twice the time to reach maximum height

2 t₁ = 3.896 s

The horizontal distance covered is

D = v × t

D = 3.896×19.09

   = 74.375 m

The ball was in the air for 3.896 seconds

The formula for the kinetic energy of any object in motion is

                           (1/2) (mass) (velocity²).

For the object you've mentioned, it translates to

                            (1/2) (100 kg) (12.5 m/s)²

                         =      (50 kg)  (156.25 m²/s²)

                         =              7,812.5 joules  
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