A small charged ball lies within the hollow of a metallic spherical shell of radius R. For three situations, the net charges on

I am unsure of the angle in your diagram; therefore, I utilized the vertical angle instead.

The new gravitational force would be calculated as follows: Consider the general formula for the gravitational force between two masses m1 and m2 that are separated by a distance "d". For the two trucks weighing 2,000 kg each, the gravitational force is initially 3.00 × 10-5 N. When one truck is loaded with an additional 1,000 kg of bricks, we need to determine the new gravitational force (x). We find this by dividing the last equation by the original to eliminate common factors (G, d^2).

It’s important to note that ‘the cart or spring is situated on a frictionless horizontal track’, which means there are no frictional forces acting on them.

Explanation:

a) Per the scenario, when the cart is drawn to position A and then let go, its initial speed at A (at time t=0) is 0 m/s. The cart travels toward position E, reaches a point where it reverses direction, and then heads back to position A. During this second phase, as the cart travels from A to E, its speed increases until it becomes zero again at point E, at which point it changes direction again, thus

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b) Let’s assume the distance between two successive points is x meters and the spring constant is k N/m

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d) Moving to the right

e) Moving to the left

Answer:

The resultant torque is zero.

Explanation:

Assuming the dipole consists of two equal but opposite charges e, it can be represented by a rod with one end featuring a charge e and the other end with -e. Since the dipole is aligned with the electric field, both charges experience forces that are parallel to this electric field. Consequently, there are no force components that act perpendicular to the rod, which is necessary for torque to occur.

Respuesta:

11.4 m/s

Explicación:

La fórmula para la aceleración centrípeta es:

donde, a es la aceleración, v la velocidad alrededor de la circunferencia y R el radio del círculo.

En este problema,

a = g = aceleración debida a la gravedad en la cima =

v = ?

R = 13.2 m

Por lo tanto,

v = 11.4 m/s