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Determine the final velocity of a proton that has an initial velocity of 2.35 x 10^5 m/s and then is accelerated uniformly in an electric field at a rate of –1.10 x 10^12 m/s/s for 1.50 x 10^-7 s.
a 7.00 x 10 ^ 3 m/s
b 7.00 x 10 ^ 4 m/s
c 7.00 x 10 ^ 5 m/s
d 7.00 x 10 ^ 6 m/s

Sagot :

lucic

Answer:

b. 7.0 x 10⁴ m/s

Explanation:

Acceleration is defined as the rate of change of velocity with time.

a= v₂-v₁ / t

a= -1.10 x 10¹² m/s²    

v₁ = 2.35 x 10⁵ m/s

v₂ = ?

t= 1.50 x 10⁻⁷ s

Applying the values to the equation

a=v₂-v₁ / t

-1.10 x 10¹² =v₂ - 2.35 x 10⁵ / 1.50 x 10⁻⁷

-1.10 x 10¹² * 1.50 x 10⁻⁷ = v₂ - 2.35 x 10⁵

-1.65 x 10⁻⁵ = v₂ - 2.35 x 10⁵

-1.65 x 10⁻⁵ + 2.35 x 10⁵ = v₂

7.0 x 10⁴ m/s = v₂

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