A modified Atwood machine is at rest. The hanging block has a mass of 3kg. The black on wheels has unknown mass M1. When release
d the block m2 falls at 5.88m/s^2. If frictional forces are considered so small they are negligible, the block M1 must have a mass of _______ kg.
1 answer:
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Answer:
t = 0.0735 m
Explanation:
Angular acceleration of the flywheel is given as
now after t = 8 s the speed of the flywheel is given as
now rotational kinetic energy of the wheel is given as
now we have
Answer:
The gravitational force is 3.509*10^17 times larger than the electrostatic force.
Explanation:
The Newton's law of universal gravitation and Coulombs law are:
Where:
G= 6.674×10^−11 N · (m/kg)2
k = 8.987×10^9 N·m2/C2
We can obtain the ratio of these forces dividing them:
--- (1)
The mass of the moon is 7.347 × 10^22 kilograms
The mass of the earth is 5.972 × 10^24 kg
And q1=q2=Na*e=(6.022*10^23)*(1.6*10^-19)C=9.635*10^4 C
Replacing these values in eq1:
Therefore
This means that the gravitational force is 3.509*10^17 times larger than the electrostatic force, when comparing the earth-moon gravitational field vs 1mol electrons - 1mol protons electrostatic field
