Two blocks of masses 1.0 kg and 2.0 kg, respectively, are pushed by a constant applied force F across a horizontal frictionless
1 answer:
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Answer:
17.71N/m
Explanation:
The period of the spring is expressed according to the expression;
m is the mass of the object
k is the force constant
Given
m = 5.50kg
T = 3.50s
Substitute into the formula;
Hence the force constant of the spring is 17.71N/m
We can apply the law of conservation of energy here. The total energy of the proton must remain constant, so the sum of the variation of electric potential energy and of kinetic energy of the proton must be zero:
which means
The variation of electric potential energy is equal to the product between the charge of the proton (q=1eV) and the potential difference (
):
Therefore, the kinetic energy gained by the proton is
<span>And since the initial kinetic energy of the proton was zero (it started from rest), then this 1000 eV corresponds to the final kinetic energy of the proton.</span>
which means
The variation of electric potential energy is equal to the product between the charge of the proton (q=1eV) and the potential difference (
Therefore, the kinetic energy gained by the proton is
<span>And since the initial kinetic energy of the proton was zero (it started from rest), then this 1000 eV corresponds to the final kinetic energy of the proton.</span>