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Basile [38]
3 years ago
5

A boy swings a ball on a string at constant speed in a circle that has a circumference equal to 6 m. What is the work done on th

e ball by the 10 N tension force in the string during one revolution of the ball

Physics
1 answer:
Grace [21]3 years ago
8 0

Answer:

0 J

Explanation:

From the diagram below; we would notice that the Force (F) = Tension (T)

Also the angle θ adjacent to the perpendicular line = 90 °

The Workdone W = F. d

W = Fd cos θ

W = Fd cos 90°

W = Fd (0)

W = 0 J

Hence the force is perpendicular to the direction of displacement and the net work done in a circular motion in one complete revolution is = 0

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Two 2.0kg bodies A and B collide The velocities before the collision are U1=15i+30j and U2=-10j+5.0j After the collision V1=-5.0
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Use the law of conservation of momentum. Since the momentum is a linear measure, we can treat each of the dimension separately:

i-direction:

m_1u_{1i}+m_2u_{2i}=m_1v_{1i}+m_2v_{2i}\\v_{2i} = \frac{m_1u_{1i}+m_2u_{2i}-m_1v_{1i}}{m_2}=\frac{(2\cdot 15-2\cdot10+2\cdot5)kg\frac{m}{s}}{2kg}=10\frac{m}{s}

j-direction:

m_1u_{1j}+m_2u_{2j}=m_1v_{1j}+m_2v_{2j}\\v_{2j} = \frac{m_1u_{1j}+m_2u_{2j}-m_1v_{1j}}{m_2}=\frac{(2\cdot 30+2\cdot5-2\cdot20)kg\frac{m}{s}}{2kg}=15\frac{m}{s}

Answer: Final velocity is: (10i + 15j) m/s

Change in the kinetic energy:

\Delta E_k = E_{ku}-E_{kv} = \frac{1}{2}m(u_1^2+u_2^2-v_1^2-v_2^2)=\\=\frac{1}{2}2kg(1125+125-425-325)\frac{m^2}{s^2}=500J

Answer: The system lost 500J worth of kinetic energy in the collision

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The deepest point in the ocean is 11 km below sea level, deeper than Mt. Everest is tall.
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