Answer:
M g H = 1/2 M v^2 potential energy = kinetic energy
v^2 = 2 g H = 2 * 9.80 * 6 = 117.6 m/s^2
v = 10.8 m/s
(C)
Answer:
c > √(2ab)
Explanation:
In this exercise we are asked to find the condition for c in such a way that the results have been real
The given equation is
½ a t² - c t + b = 0
we can see that this is a quadratic equation whose solution is
t = [c ±√(c² - 4 (½ a) b)] / 2
for the results to be real, the square root must be real, so the radicand must be greater than zero
c² -2a b > 0
c > √(2ab)
Ideal M.A. is 1 I.e, load =effort
The minimum initial velocity that the ball must have for it to reach the top of the hill is 21 m/s. The correct option is D.
<h3>What is mechanical energy?</h3>
The mechanical energy is the sum of kinetic energy and the potential energy of an object at any instant of time.
M.E = KE +PE
A boy is trying to roll a bowling ball up a hill. The friction is ignored. The ball must have to reach the top of the hill with a velocity. The acceleration due to gravity, g = 9.8 m/s²
The conservation of energy principle states that total mechanical energy remains conserved in all situations where there is no external force acting on the system.
M.E bottom of hill = M.E on top of hill
Kinetic energy + Potential energy = Kinetic energy + Potential energy
1/2 mu² + 0 = 0 + mgh
At the top of hill, the velocity will become zero. So, final kinetic energy is zero.
Substituting the values, we have
1/2 x u² = 9.8 x 22.5
u = sqrt [2 x9.8 x 22.5 ]
u= 21 m/s
Thus, the minimum initial velocity that the ball must have for it to reach the top of the hill is 21 m/s.
Learn more about mechanical energy.
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