Electric field due to a point charge is given as

here we know that

also the distance is given as

now we will have

so we will have

so above is the electric field due to proton
Answer:
Option D
490 J
Explanation:
When at a height of 100 am above and released, the ball initially posses only potential energy. When it falls, some potential energy is converted to kinetic energy.
Initial potential energy= mgh where m is the mass, g is the acceleration due to gravity and h is height. Substituting 1 Kg for m, 9.81 for g and 100 m for h then
PE initial = 1*9.81*100= 981 J
At 50 m, PE will be 1*9.81*50=490.5 J
Subtracting PE at 50 m from initial PE we get the energy that has been converted to kinetic energy hence
981-490.5= 490.5 J
Approximately, 490 J
Answer:
The duration of the impact is 0.005384 seconds
Explanation:
Given
m = 0.43 kg
v = 5.2 m/s
x = 0.014 m
Knowing the formulas

It depends on Mass and velocity
Answer:
Data:-vi=om/s (b/c as in question penny is dropped from building means before coming to ground its initial state or velocity was considered as zero ) now distance or height h=380m and now we have to find the final velocity vf=? and the time t=?
Explanation:
So applying second eq of motion s=vit+1/2×gt² (here we have taken a gravity b/c when ever body is in vertical position then acceleration due to gravity is applied ) s=0×t+1/2×gt² , s=0+1/2×9.8×t² ,380=4.9t² we have to find t so 4.9t²=380 , t²=380÷4.9 , t²=77.55 now sq root on b/s

so t=8.806s and now apply 1st eq o²f motion to find out vf so vf=vi+gt , vf=0+9.8×8.806 ,vf=86.298 and if you want to verify that either this is answer is correct or not so put the value of t in second eq of motion and if you got distance same as give in the question so your value of t is considered as correct likewise s=vit+1/2gt² , s=0+1/2×9.8(8.806)²,s=4.9×77.55 ,s=380m (proved) I hope it would be helpfull