D. Using a fixed-pulley system
Answer:
4.45×10¯¹¹ N
Explanation:
From the question given above, the following data were obtained:
Mass of ball (M₁) = 4 Kg
Mass of bowling pin (M₂) = 1.5 Kg
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Distance apart (r) = 3 m
Force of attraction (F) =?
The force of attraction between the ball and the bowling pin can be obtained as follow:
F = GM₁M₂ / r²
F = 6.67×10¯¹¹ × 4 × 1.5 / 3²
F = 4.002×10¯¹⁰ / 9
F = 4.45×10¯¹¹ N
Therefore, the force of attraction between the ball and the bowling pin is 4.45×10¯¹¹ N
We have that the values for F north,
F east,
F up are
From the Question we are told that
electric force 
electric force , 
electric force , 
charge on this ball one 
charge on this ball two 
Generally the equation for the F north is mathematically given as


For F East


For F UP


For more information on this visit
brainly.com/question/21811998
Scientific Notation: 4.580 x 10^-4
Scientific e Notation: 4.580e-4
energy is the correct answer to fill the blank bb :)