Rey lifts a 6,300 g metal ball from the ground to a height of 98.15 cm close to his body. (a) What is the balls PEg? Realizing that the ball is heavy, he suddenly releases it with a speed of 15m/sa. (b) what is the balls KE? Given:
m= 6,300 g =6.3 kg
h= 98.15 cm =0.9815 m

Formula:
a) PE= mgh
PE=
PE=

[v= 15 m/s]
b) KE= mv²/2
KE=
KE=
​

B) erini just took the test and it is correct

Answer: a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same sign as the  -4.0  µc charge.======electricstatic force between two point charges  is given by    where k is coulomb's constant,  8.99×10⁹  n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ =  6.0 µc, q₂ = -4.0  µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1  µc = 10⁻⁶ c)    the force from q₂    the force from q₃    the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃    isolating:     a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same as the  -4.0  µc charge