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icang [17]
4 years ago
10

As an object is undergoing free fall motion.As it falls the objects?,speed increases,acceleration increases,both of these,or non

e
Physics
1 answer:
Alex_Xolod [135]4 years ago
4 0
The speed will increase, as there is acceleration, while the acceleration will remain constant, as gravity is constant.
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a compound machine is made of several simple machines

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2 years ago
In a ballistic pendulum, a spring pushes a ball from rest It ies through the air and sticks in the base of a pendlum that swings
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Answer:

The final velocity is \bf{562.9}.

Explanation:

Given:

The maximum angle that the pendulum and the ball system can travel, \theta_{m} = 45^{0}.

The length of the pendulum, l = 30~cm.

The mass of the pendulum, m_{p} = 250~g.

The mass of the ball, m_{b} = 76~g.

Consider that the initial velocity of the ball-pendulum system is v_{i}. So the initial total energy of the system is given by

E_{i} = K.E. + P.E.\\~~~~= \dfrac{1}{2}(m_{b} + m_{s})v_{i}^{2} + 0\\~~~~= \dfrac{1}{2}(m_{b} + m_{s})v_{i}^{2}

Consider the final velocity of the ball is v_{0} and the final height attended by the system is h. So the final total final energy of the system is given by

E_{f} = K.E. + P.E.\\~~~~~= 0 + (m_{b} + m_{s})gh\\~~~~~= (m_{b} + m_{s})gh

From the conservation of energy,

E_{i} = E_{f}\\&or,& \dfrac{1}{2}(m_{b} + m_{s})v_{i}^{2} = (m_{b} + m_{s})gh\\&or,& v_{i} = \sqrt{2gh}

From the conservation of momentum,

&& m_{b}v_{0} = (m_{b} + m_{s})v_{i}\\&or,& v_{0} =  (1 + \dfrac{m_{p}}{m_{b}})\sqrt{2gh}

The final height attended by the system is given by

h = l(1 - \cos 45^{0}) = 8.787~cm

The final velocity is given by

v_{0} = (1 + \dfrac{250}{76}})\sqrt{2(980)(8.787)}\\~~~~= 562.9

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