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kherson [118]
3 years ago
14

A coin mass 0.005 kg is dropped from a height of 3 m. How much mechanical energy does it have right before it hits the ground ?

Physics
1 answer:
lesya692 [45]3 years ago
6 0
Mechanical energy equals the sum of potential and kinetic energy. During the process, all PE converts into KE, assuming air resistance is neglected. So, the mechanical energy does not change and is equal to the initial potential energy.
ME
=mgh
=0.005 x 9.81 x 3
=0.147J
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Answer:

\frac{d^2x}{dt^2}+\frac{\beta}{m}\frac{dx}{dt}+\frac{k}{m}x=0

Explanation:

let m be the mass attached, let k be the spring constant and let \beta be the positive damping constant.

-By Newton's second law:

m\frac{d^2x}{dt^2}=-kx-\beta \frac{dx}{dt}

where x(t) is the displacement from equilibrium position. The equation can be transformed into:

\frac{d^2x}{dt^2}+\frac{\beta}{m}\frac{dx}{dt}+\frac{k}{m}x=0  shich is the equation of motion.

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a_{radial} = \frac{v^2}{r}

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A 10.2-kg mass is located at the origin, and a 4.6-kg mass is located at x = 8.1 cm. Assuming g is constant, what is the locatio
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Answer:

center of mass of the two masses will lie at x = 2.52 cm

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Explanation:

Position of centre of mass is given as

r_{cm} = \frac{m_1r_1 + m_2r_2}{m_1 + m_2}

here we have

m_1 = 10.2 kg

m_2 = 4.6 kg

r_1 = (0, 0)

r_2 = (8.1cm, 0)

now we have

r_{cm} = \frac{10.2 (0,0) + 4.6 (8.1 , 0)}{10.2 + 4.6}

r_{cm} = {(37.26, 0)}{14.8}

r_{cm} = (2.52 cm, 0)

so center of mass of the two masses will lie at x = 2.52 cm

now for center of gravity we can use

r_g_{cm} = \frac{m_1gr_1 + m_2gr_2}{m_1g + m_2g}

here we have

m_1 = 10.2 kg

m_2 = 4.6 kg

r_1 = (0, 0)

r_2 = (8.1cm, 0)

now we have

r_g_{cm} = \frac{10.2(9.8) (0,0) + 4.6(9.8) (8.1 , 0)}{10.2(9.8) + 4.6(9.8)}

r_g_{cm} = {(37.26, 0)}{14.8}

r_g_{cm} = (2.52 cm, 0)

So center of mass is same as center of gravity because value of gravity is constant here

3 0
3 years ago
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