(please help!) A 60Kg crate is pulled with a force of 400N at an angle of 30 degrees above the horizontal. A 150N force resists sliding of the crate. (Be sure to do a sketch (FBD) A) Calculate the weight of the crate:
B) Calculate the components of the 400N force: Fx = . Fy =
C) Calculate the normal force on the crate:
D) Calculate the acceleration of the crate:
E) Calculate the speed of the crate if it started from rest after t = 4 s

Hmmm lemme come back in 20 minutes with an answer

Indoraptor for sure

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