Answer:
because you rubbed the ballon the charge went faster
Explanation:
Answer:
3360 N
Explanation:
In a first-class lever, the effort force and load force are on opposite sides of the fulcrum.
The lever is 5 m long. The load force is 1.50 m from the fulcrum, so the effort force must be 3.50 m from the fulcrum.
The torques are equal:
Fr = Fr
(1440 N) (3.5 m) = F (1.5 m)
F = 3360 N
When an object is thrown upwards, it reaches a maximum height and then comes back to its starting position. If we put f(t) = 0, we can get the time the ball was thrown and the time it returned.
Solving:
-16t² + 94t + 12 = 0
t = -0.125 and t = 6; negative time is not possible so the first solution is discarded.
The maximum height lies at the time exactly between the two times of zero displacement. Max height at
t = (-0.125 + 6)/2 = 2.9375 s ≈ 3
3 < t < 6
Therefore, the answer is the first option.
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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.