A body of mass 2.9 kg is suspended from a string of length 2.5 metres and is at rest. A bullet of mass 100 gram moving horizonta
lly with a speed of 150 metres per second strikes and sticks to it. what is the maximum angle made by the string with vertical after the impact.
(g=10m/s)
(g=10m/s)
1 answer:
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Answer:
The acceleration of the proton is 2.823 x 10¹⁷ m/s²
The acceleration of the electron is 5.175 x 10²⁰ m/s²
Explanation:
Given;
distance between the electron and proton, r = 7 x 10⁻¹⁰ m
mass of proton, = 1.67 x 10⁻²⁷ kg
mass of electron, = 9.11 x 10⁻³¹ kg
The attractive force between the two charges is given by Coulomb's law;
where;
k is Coulomb's constant = 9 x 10⁹ Nm²/c²
Acceleration of proton is given by;
F = ma
Acceleration of the electron is given by;