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rusak2
3 months ago
7

A very small round ball is located near a large solid sphere of uniform density. The force that the large sphere exerts on the b

all A very small round ball is located near a large solid sphere of uniform density. The force that the large sphere exerts on the ball a. is independent of the mass of the sphere. b. can only be calculated using calculus. c. is approximately the same as it would be if all the mass of the sphere were concentrated at the center of the sphere. d. is independent of the mass of the ball. e. is exactly the same as it would be if all the mass of the sphere were concentrated at the center of the sphere.
Physics
1 answer:
Softa [3K]3 months ago
5 0

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.

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Yuliya22 [3333]

Answer: Her velocity magnitude (v) relative to the shore is 5.70 km/h.

Explanation:

Let Q be the speed of the boat, and P be the speed of the river flow.

R represents the resultant velocity combining boat velocity and river current.

According to vector addition using the law of triangles:

R=\sqrt{P^2+Q^2+2PQCos\theta}

From the diagram:

P = 3.5 km/h, Q = 4.5 km/h

\theta= 90^o

R=\sqrt{P^2+Q^2+2PQCos\theta}=\sqrt{(3.5)^2+(4.5)^2+3.5\times 4.5\times cos90^o}=5.70

(Cos90^o=0),(sin 90^o=1)

\alpha =tan^{-1}\frac{Qsin\theta}{P+Qcos\theta}=tan^{-1}\frac{4.5 sin 90^o}{3.5+4.5 cos90^o}=tan^{-1}\frac{4.5}{3.5}=52.12^o

Therefore, her velocity magnitude relative to the shore is 5.70 km/h.

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3 months ago
Several charges in the neighborhood of point P produce an electric potential of 6.0 kV (relative to zero at infinity) and an ele
ValentinkaMS [3465]

Answer:

0.018 J

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The work required to bring the charge from infinity to the point P is equal to the change in its electric potential energy. This can be expressed as

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After substituting into the formula, we arrive at

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A 5.8 × 104-watt elevator motor can lift a total weight of 2.1 × 104 newtons with a maximum constant speed of
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A 1000-kg car is moving along a straight road down a 30∘30∘ slope at a constant speed of 20.0m/s20.0m/s. What is the net force a
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Let's evaluate the situation separately for the vertical direction and the horizontal direction along the slope.

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Learn more about slopes and friction:

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