A thin uniform-density rod whose mass is 3.0 kg and whose length is 2.6 m rotates around an axis perpendicular to the rod, with angular speed 37 radians/s. Its center moves with a speed of 13 m/s. (a) What is its rotational kinetic energy

rotational kinetic energy is 1156.81 J

Explanation:

Given the data in the question;

first we find the moment of inertia of the rod as axis passes through the middle or center;

where m is the mass of the road( 3.0 kg)

is the length of the rod ( 2.6 m )

so we substitute

× 3 × (2.6)²

=   1.69 kg.m²

now, to calculate the rotational kinetic energy,

kinetic energy due to rotation is;

KE =

where is moment of inertia( 1.69 kg.m² ),

w is angular frequency ( 37 radians/s )

we substitute

KE = × 1.69 × ( 37 radians/s )²

KE = 1156.81 J

Therefore, rotational kinetic energy is 1156.81 J

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.