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LuckyWell [14K]
3 years ago
12

These two pls :)))) ill mark brainliest :)

Physics
1 answer:
Anna71 [15]3 years ago
4 0

Answer:

Bowling Ball: weight on Earth = 49 N

Textbook: Mass = 2 kg; weight on the moon = 3.2 N

Large dog: weight on Earth = 490 N; weight on the moon = 80 N

Law of Universal Gravitation: F_{G}=\frac{Gm_{1}m_{2}}{r^{2}}

F_{G} = gravitational force (Newtons/N)

<em>G</em> = gravitational constant, 6.67430 × 10¹¹ \frac{N*m^{2}}{kg^{2}}

<em>m</em>₁ and <em>m</em>₂ = masses of two objects (kilograms/kg)

<em>r</em>² = square of distance between centers of the two objects (meters/m)

Have a fantastic day!

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A 86g ball is dropped vertically to the floor from a height of 2.87m and bounces to a height of 1.28. What is the magnitude of t
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Answer:

The impulse received by the ball from the floor during the bounce is approximately 1.11329438 m·kg/s

Explanation:

The given mass of the ball, m = 86 g = 0.089 kg

The height from which the ball is dropped, H = 2.87 m

The height to which the ball bounces, h = 1.28 m

Mathematically, we have;

Δp = F·Δt

Where;

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The velocity with which the ball leaves, v₂ = √(2·g·h)

The change in momentum, Δp = m·(v₂ - v₁)

∴ Δp = m·(√(2·g·h) - (-√(2·g·H))) = m·(√(2·g·h) +√(2·g·H) )

The impulse, Δp, received by the ball from the floor during the bounce is given as follows;

Δp = 0.089 kg × (√(2 × 9.8 m/s² × 1.28 m) + √(2 × 9.8 m/s² × 2.87 m)) ≈ 1.11329438 m·kg/s

The impulse received by the ball from the floor during the bounce, Δp ≈ 1.11329438 m·kg/s

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