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Lemur [1.5K]
4 years ago
6

Two balls of equal mass at the bottom of an incline are rolled upward without slipping at the same initial velocity. one ball is

solid and the other is a thin-walled hollow ball. which rolls higher up the incline before coming to a stop?
Physics
1 answer:
german4 years ago
3 0
The appropriate response would be the hollow ball.The purpose for it has more kinetic energy at the last (2/3 as opposed to 2/5) so it will have more potential energy (more height) at the best. Pondering it, more inertia implies its less impervious to change meaning the hollow ball with more inertia wont need to stop as fast.
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Three uniform spheres are located at the corners of an equilateral triangle. each side of the triangle has a length of 1.20 m. t
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Refer to the diagram shown below.

Because of symmetry, equal forces, F, exist between the sphere of mass m and each of the other two spheres.
The acceleration of the sphere with mass m will be vertical as shown.

The gravitational constant is G = 6.67408 x 10⁻¹¹ m³/(kg-s²)

Calculate F.
F = [ (6.67408 x 10⁻¹¹ m³/(kg-s²))*(m kg)*(2.8 kg)]/(1.2 m)²
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The resultant force acting on mass m is
2Fcos(30°) = 2*(1.2977 x 10⁻¹⁰m N)*cos(30°) = 2.2477 x 10⁻¹⁰m  N

If the initial acceleration of mass m is a m/s², then
(m kg)(a m/s²) = (2.2477 x 10⁻¹⁰m N)
a = 2.2477 x 10⁻¹⁰ m/s²

Answer:
The magnitude of the acceleration on mass m is 2.25 x 10⁻¹⁰ m/s².
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7 0
3 years ago
A 350-g air track cart is traveling at 1.25 m/s and a 280-g cart traveling in the opposite direction at 1.33 m/s. What is the sp
kumpel [21]

Answer:

The speed of the center of mass of the two carts is 0.103 m/s

Explanation:

It is given that,

Mass of the air track cart, m₁ = 350 g = 0.35 kg

Velocity of air track cart, v₁ = 1.25 m/s

Mass of cart, m₂ = 280 g = 0.28 kg

Velocity of cart, v₂ = -1.33 m/s (it is travelling in opposite direction)

We need to find the speed of the center of mass of the two carts. It is given by the following relation as :

v_{cm}=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}

v_{cm}=\dfrac{0.35\ kg\times 1.25\ m/s+0.28\ kg\times (-1.33\ m/s)}{0.35\ kg+0.28\ kg}

v_{cm}=0.103\ m/s

Hence, this is the required solution.

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