on planet tehar, the free fall acceleration is the same as that on earth, but th

Question

on planet tehar, the free fall acceleration is the same as that on earth, but there is also a strong downward electric field that is uniform close to the planet's surface. a2.00-kg ball having a charge of 5.00 micro C is thrown upward at a speed of 20.1 m/s. it hits the ground after an interval of 4.10 s. What is the potential difference between the starting point and the top point of the trajectory?

Explanation / Answer

After reaching initial point projrction Acceleration(a) = 2 v / t = ( 2 * 20.1) / 4.10 = 9.8048 m/s2 --------------(1) Net force downward on the plant is : F = Eq + mg HenceAcceleration (a) = ( Eq + mg) /m ----------------(2) From equation (1) and (2) ; ( Eq + mg) / m = 9.8048 ==> Eq /m = 9.8048 - 9.8 = 0.0048 ==> E = 0.0048 * m / q = (0.0048 * 2.0) / (5.0 *10-6 ) = 1935.32 N/C We know that : Potentialdifference is : V = E * (? Y) -----------(3) Here ? Y = v t - 0.5 at2 ; Here t = 5.01/2 = 2.505 s = (20.1 * 2.505) - 0.5 * 9.8048 * (2.505)2 = 19.5878 m Hence V = E * (? Y) = 1935.32 N /C * 19.5878 m = ------------- V

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.