Explanation:
<em><u>Newton's First Law ,</u></em>
F = ma
a = m/F
a = 68 / 59
a = 1.15 m/s2
Potential energy = m · g · h
-- When you held the ball at 2.0 meters above the floor, it had
(0.5 kg) · (9.8 m/s²) · (2.0 m) = 9.8 Joules of potential energy.
-- After it bounced and went back up as high as it could, it was only 1.8 meters above the floor. Its potential energy was
(0.5 kg) · (9.8 m/s²) · (1.8 m) = 8.82 Joules
-- Between the drop and the top of the bounce, it lost
(9.8 - 8.82) = <em>0.98 Joule</em> .
-- The energy was lost when the ball hit the floor. During the hit, 0.98 joule of kinetic energy turned to <em>thermal energy</em>, which slightly heated the ball and the floor.
Answer:
Torque on the coil will be ZERO
Explanation:
As we know that the magnetic moment of the closed current carrying coil is always along its axis and it is given as
now we know that magnetic field is also along the axis of the coil so here as we know the equation of torque given as
so we have
The log of 4311 is 8.3689251747471
