Two cars are heading towards each other but are 12 km apart. one car is going 70 km/hr, and the other is going 50 km/hr. how muc
1 answer:
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Answer:
0.34 sec
Explanation:
Low point of spring ( length of stretched spring ) = 5.8 cm
midpoint of spring = 5.8 / 2 = 2.9 cm
Determine the oscillation period
at equilibrum condition
Kx = Mg
g= 9.8 m/s^2
x = 2.9 * 10^-2 m
k / m = 9.8 / ( 2.9 * 10^-2 ) = 337.93
note : w = =
= 18.38 rad/sec
Period of oscillation =
= 0.34 sec
Answer:
<em>The kinetic energy of a spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.</em>
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Explanation:
Let us first consider the initial characteristics of the angular motion of the disk
moment of inertia =
angular speed = ω
For the second case, we consider the characteristics to now be
moment of inertia = (five times larger)
angular speed = ω/5 (five times smaller)
Recall that the kinetic energy of a spinning body is given as
therefore,
for the first case, the K.E. is given as
and for the second case, the K.E. is given as
<em>this is one-tenth the kinetic energy before its spinning characteristics were changed.</em>
<em>This implies that the kinetic energy of the spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.</em>