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Ksenya-84 [330]
3 years ago
14

At the 1 s mark, which objects have equal momentum?

Physics
1 answer:
Rudik [331]3 years ago
3 0
C. The bowling ball and the bicycle

p = mv
P of bike = 12x5 = 60
P of rock = 2x20 = 40
P of ball = 5 x 10 = 60
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B. people with OCD know their disorder is irrational

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What is the distance from axis about which a uniform, balsa-wood sphere will have the same moment of inertia as does a thin-wall
andrey2020 [161]

Answer:

D_{s} ≈ 2.1 R

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The moment of inertia of the bodies can be calculated by the equation

     I = ∫ r² dm

For bodies with symmetry this tabulated, the moment of inertia of the center of mass

Sphere               Is_{cm} = 2/5 M R²

Spherical shell   Ic_{cm} = 2/3 M R²

The parallel axes theorem allows us to calculate the moment of inertia with respect to different axes, without knowing the moment of inertia of the center of mass

    I = I_{cm} + M D²

Where M is the mass of the body and D is the distance from the center of mass to the axis of rotation

Let's start with the spherical shell, axis is along a diameter

     D = 2R

    Ic = Ic_{cm} + M D²

    Ic = 2/3 MR² + M (2R)²

    Ic = M R² (2/3 + 4)

    Ic = 14/3 M R²

The sphere

    Is =Is_{cm} + M [D_{s}²

    Is = Ic

    2/5 MR² + M D_{s}² = 14/3 MR²

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3 0
3 years ago
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
irga5000 [103]

Answer:

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

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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;

Δp = The change in momentum = m·Δv

F = The applied force

Δt = The time of contact with the force

The velocity of the ball just before it touches the ground, v₁ = -√(2·g·H)

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

6 0
3 years ago
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