A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s2 starts from rest at t = 0. The wheel has a diameter of 20 cm. What is the magnitude of the total linear acceleration of a point on the outer edge of the wheel at t = 0.60 s? Hint: To find total linear acceleration (in m/s2)you need to have tangential and radial acceleration . Tangential acceleration can be calculated using angular acceleration at=rα To find redial acceleration you need to calculate final linear speed of that point and use ar=vf2r . And vf=rωf.

He can play two games. 13/4=3.253.25*2=6.5

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