Question

C UTIONS TO THE STATE T OUCO C RES positive sodium lon (Na) from the interior of the cell? 0000000000000000176 Confirm Mac Password To continue using cloud, confirm you View Ann. Update Apple ID Settings Some con services will not be available in you sign in again View 3. - 10 points SerCP10 16.P.010.MI. B My Notes On planet ear, the free-fall acceleration is the same as that on Earth, but there also a strong downward electric field that is uniform close to the planet's surface. A 1.82-kg ball having a charge of 4.64 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? (Use 9.8 m/s for the acceleration due to gravity) 4. - 10 points SerCP10 16.P.013. My Notes Consider the following figure. 8.00 C 200 C 400C (a) Find the electric potential, taking rere at Infinity, at the upper right corner (the corner without a charge of the rectangle in the figure. (Let x 5.00 cm and y 2.20 cm.) (6) Repeat if the 2.00-C charge is replaced with a charge of -2.00 C. 10.10.2022 Mi Notes

Answer 1

Let the electric field-E charge (q) = 4.64 uc= 4.64 x 10 °C mass cm) = 1.92 kq upward speed cu) = 20.1 m/s time of flight = 4.108 time taken to reach the highest acceleration due to field = 8E Net acceleration = 9LE Point (t) = 4.10 2 22-058 m NOLD + a 8 + 8 - m = E= - 2011 2.05 1.9134x10 % uz= (20.172 2x9.804 878815 height cho= so Potential v= - 2o. 6075 m difference hE= 3.9420894 x 10 W No-y 139.420520kv 8. ELC Herce x= 5cm=5x16?m y = 2.2 x 102m BD = FR 2+ y2 - 5.463x10 m 2.00 uc 4.colle @ ve une [ Bot] *to* axiol 2009 10 2 + + 2 5. 4 3 = 340.58534) 1 2.2 V (6) FOTC -2 lle 9x109.06 VE an [ t 4 2.2 #] [v=2*4ucay V.274.4690 5 5.463