Answer:


Explanation:
k = Coulomb constant = 
Q = Charge
r = Distance = 8 cm
R = Radius = 4 cm
Electric field is given by

Volume charge density is given by

The volume charge density for the sphere is 

The magnitude of the electric field is 
Because of symmetry electric field component in the x axis cancels out. Now just use electric field formula and slap that sine of theta cause you want the vertical component of electric field and multiply that by two since there’s two charges. I’ve shown my work. Hope it helps✌
Answer:
the answer is that the dough has the same mass before and after it was flattened
Answer:

Explanation:
v = Velocidad final = 
u = Velocidad inicial = 0
t = Tiempo empleado = 15 s
a = Aceleración
De las ecuaciones cinemáticas tenemos

La aceleración del camión en el primer intervalo de tiempo es
.
Ill save you all the math steps, but here is the answer! <span>102.25m I took that physics exam 3 days ago! So if you need the steps just ask Ill insert them in!</span>