Heeeppp! i will mark brianlist show your work

- with the device starting from a still position (0 m/s), you drop it, allowing gravity to accelerate it towards the ground at 9.8 m/s2. assuming it takes 1.5 seconds for the device to hit the ground, what is it’s velocity? [v = a*t]

- momentum is defined as mass * velocity [p = m*v]. calculate the momentum of the device just before it hits the ground using the velocity you calculated in the last question.

- the equation pe = m*g* using h, what is the potential energy (pe) of the device before it is dropped? [remember, h = height, m = mass, g = acceleration due to gravity]

- what is the kinetic energy that the device contains immediately before it hits the ground? [ke = ½*m*v2]

Initial velocity,  u = 4.9 m/sfinal velocity = v m/sinitial height,  h₁ =  245 m    final height, h₂ = 0a = -9.8  m/s²time taken = t secondss = ut + 0.5at²⇒  (h₂-h₁) = ut + 0.5at²⇒  0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get   t = 7.59sv  = u +  at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is  69.48 m/s downward

Answer (d) valerie rides her bike south at 8 kilometers per hour///owen jogs south at 8 kilometers per hour ;

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