The magnitude of the net force versus time graph has a rectangular shape. Often in physics geometric properties of graphs have p

Respuesta:

Opción e

Explicación:

La Ley de Gravitación Universal indica que toda masa puntual atrae a otra masa puntual en el universo con una fuerza que se dirige en línea recta entre los centros de masa de ambos, siendo esta fuerza proporcional a las masas de los objetos y inversamente proporcional a su separación. Esta fuerza atractiva siempre es dirigida del uno hacia el otro. La ley es aplicable a objetos de cualquier masa, sin importar su tamaño. Dos objetos grandes pueden ser considerados masas puntuales si la distancia entre ellos es considerablemente mayor que sus dimensiones o si presentan simetría esférica. En tales casos, la masa de cada objeto puede ser modelada como una masa puntual en su centro de masa.

La misma fuerza actúa sobre ambas bolas.

Since it's classified as a transverse wave, the particle on the string moves horizontally as the wave progresses, without actual forward or backward travel. Consequently, the red dot shifts 'A' to the left, returns 'A' to the center, moves 'A' to the right, and goes back 'A' to the center once again. Thus, the red dot collectively travels a distance totaling 4A.

<span>Response: Chlorine has 17 electrons, thus, for 1+ and 2+ ions, we require elements with 18 and 19 electrons, which are argon and potassium: Ar+ and K 2+. For 1- and 2- ions, we need elements with 16 and 15 electrons, namely sulfur and phosphorus, represented as S- and P 2-. It’s important to note that + ions indicate electron loss, while - ions reflect electron gain.</span>

Answer:

The tension in the string when the speed increased is 134.53 N

Explanation:

Given;

Tension in the string, T = 120 N

initial speed of the transverse wave, v₁ = 170 m/s

final speed of the transverse wave, v₂ = 180 m/s

The wave speed is expressed as;

where;

μ represents mass per unit length

The new tension T₂ will be computed as;

Consequently, the tension in the string when the speed was increased is 134.53 N

a)

b) 803.4 N

Explanation:

a) At point C (top of the loop), the pilot experiences weightlessness, leading to the normal force from the seat being zero:

N = 0

Consequently, the force balance equation at position C becomes:

where the left term signifies the pilot's weight and the right term represents the centripetal force, with:

= acceleration due to gravity

= jet's velocity at the top

= loop radius

By solving for v,

Thus, this is the jet's speed at C.

The speed at position A (bottom) can be derived from

The distance traveled by the jet corresponds to half the circumference of the circle with radius r, therefore

Given the plane's deceleration is consistent, we can obtain it using the following equation:

b) The pilot experiences a force equal to the normal force from the seat. At point B (half-way through the loop), we find:

- The normal force from the seat, N, directed towards the center of the loop

- As there are no further forces acting toward the central axis, N must equal the centripetal force:

(1)

where

represents the speed at position B.

To deduce the velocity at B, we note that the distance covered by the jet between positions A and B is a quarter of a circle:

With knowledge of the deceleration, we can implement the equation of motion to find the velocity at the midway point B:

Thus, we then apply eq.(1) to determine the normal force acting on the pilot at B: