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Sagot :
When a proton slows from 0. 8c up to 0. 6c, it’s kinetic energy decreases by what factor of 1/4.
Kinetic energy of a proton having a charge Q and Voltage V is given by:
Kinetic energy = Q×V
Where, Q is the charge
V is the Voltage
Kinetic energy is directly proportional to the charge on a proton.
Let the kinetic energy of proton having charge 0.8C be KE₁ and kinetic energy of proton having charge 0.6C be KE₂
Hence KE₁ = 0.8 × V
KE₂ = 0.6 × V
Now, KE₁ / KE₂ = 0.8 V / 0.6 V
⇒ KE₁ / KE₂ = 4 / 3
⇒ KE₂ / KE₁ = 3/4
⇒ KE₂ = 3/4 times KE₁
Hence, the final kinetic energy becomes 3/4 times the initial kinetic energy.
The kinetic energy decreases by a factor of:
Change = KE₁ - KE₂
Change = KE₁ - 3/4KE₁
Change = 1/4 KE₁
Hence, the kinetic energy decreases by a factor of 1/4
Learn more about Kinetic energy here, https://brainly.com/question/12669551
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