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2 years ago

14

Physics

1 answer:

romanna [79]2 years ago

6 0

Answer: 0.00068 N

Explanation: Universal gravitational constant=6.674 *10^(-11)

Force=Gm1m2/(r^2)

Force= 6.67*25000*40000*10^(-11)/(10^2)

Force=0.00068 N

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Read 2 more answers

A hollow cylinder that is rolling without slipping is given a velocity of 5.0 m/s and rolls up an incline to a vertical height o

inysia [295]

Answer:

The hollow cylinder rolled up the inclined plane by 1.91 m

Explanation:

From the principle of conservation of mechanical energy, total kinetic energy = total potential energy

$M.E_T = \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2 + mgh$

The total energy at the bottom of the inclined plane = total energy at the top of the inclined plane.

$\frac{1}{2}mv_i^2 + \frac{1}{2} I \omega_i^2 + mg(0) = \frac{1}{2}mv_f^2 + \frac{1}{2} I \omega_f^2 + mgh$

moment of inertia, I, of a hollow cylinder = ¹/₂mr²

substitute for I in the equation above;

$\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}mr^2 \omega_i^2) = \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}mr^2 \omega_f^2) + mgh\\\\ but \ v = r \omega\\\\\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}m v_i^2 ) = \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}m v_f^2) + mgh\\\\\frac{1}{2}mv_i^2 +\frac{1}{4}mv_i^2 = \frac{1}{2}mv_f^2 +\frac{1}{4}mv_f^2 +mgh\\\\\frac{3}{4}mv_i^2 = \frac{3}{4}mv_f^2 +mgh\\\\mgh = \frac{3}{4}mv_i^2 - \frac{3}{4}mv_f^2\\\\gh = \frac{3}{4}v_i^2 - \frac{3}{4}v_f^2\\\\$

$h = \frac{3}{4g}(v_1^2 -v_f^2)$

given;

v₁ = 5.0 m/s

vf = 0

g = 9.8 m/s²

$h = \frac{3}{4g}(v_1^2 -v_f^2) =\frac{3}{4*9.8}(5^2 -0) = 1.91 \ m$

Therefore, the hollow cylinder rolled up the inclined plane by 1.91 m

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2 years ago