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3 years ago

The force of attraction between a ball is F=.........×10^-¹¹

Physics

1 answer:

DIA [1.3K]3 years ago

8 0

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

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A photon of red light (wavelength = 710 nm) and a ping-pong ball (mass = 2. 40 × 10^-3 kg) have the same momentum. At what speed

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$pball = pphotons = \frac{6.626 \times 10 {}^{ - 34}kgm/s }{700 \times 10 {}^{ - 9} }$

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learn more about momentum of photon from here: brainly.com/question/28197406

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