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Mekhanik [1.2K]
3 years ago
7

a force 10N drags a mass 10 kg on a horizontal table with acceleration 0.2m\s. If the acceleration due to gravity is 10m\s2, a c

oefficient of friction between the moving mass and a table is?​
Physics
1 answer:
kakasveta [241]3 years ago
7 0

Answer:

Explanation: 0.02

Recall: μ mg = ma

μ = ?

m = 10N

g = 10m\s2

a = 0.2m\s

Substituting the values , we have

μ x 10 x 10 = 10 x 0.2

100  μ = 2

μ = 2/100

μ = 1/50

μ = 0.02

μ = 0.02

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Mass of Neutron star = 5.5 x 10¹² kg

Mass of star Betelgeuse = 2.188 x 10³¹ kg

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\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.765 \times 10^{30}}{2\times 10^{30}} \times [\frac{696,340,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 13,675.86

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