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The total energy TE = mgh + 1/2 mU^2; where h = 20 m, g = 9.81 m/sec^2, and U = 10 mps. When the ball reaches max height H, all that TE will be potential energy PE = mgH = TE.
So there you are. TE = mgh + 1/2 mU^2 = mgH = TE from the conservation of energy. Solve for H.
1) H = (gh + 1/2 U^2)/g = h + U^2/2g = ? meters where everything on the RHS is given. You can do the math.
2) As the ball drops from H to h, it picks up KE as the potential energy mgH is converted when the potential energy is diminished to mgh, where h < H. So PE - pe = ke = mg(H - h) = 1/2 mv^2 so solve for v = sqrt(2g(H - h)) and, again, everything is given. You can do the math.
3) Same deal as 2) except now its V = sqrt(2gH) because all the PE = mgH = 1/2 mV^2 = KE when it is about to hit the ground. You can do the math.
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11 m/s simply divide the momentum with the mass
So we know F=ma
m means mass and a means acceleration
so Force= ma
so F=1300X1.07=1391N
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<span>Calcium fluoride is produced. It is one Calcium atom bonded to two fluoride atoms. The Calcium atom receives one electron from one fluoride and a second electron from the second fluoride atom. As a result of receiving the two electrons, Calcium will have a negative charge. The two fluorides become positive from losing an electron. The positive and negative charge attraction keeps the atoms close together to form an ionic bond.</span>
