It’s true the acceleration of falling objects on earth due to gravity is 98ms2
It can be both flat or it can be when you have new eyeglasses on and you look down it makes you think the ground looks like that but its not
Sprry o cant see the words clearly
<span>We can assume that the horizontal surface has no friction and the pulley is massless. We can use Newton's second law to set up an equation.
F = Ma
F is the net force
M is the total mass of the system
a is the acceleration
a = F / M
a = (mb)(g) / (ma + mb)
a = (6.0 kg)(9.80 m/s^2) / (6.0 kg + 14.0 kg)
a = 58.8 N / 20 kg
a = 2.94 m/s^2
The magnitude of the acceleration of the system is 2.94 m/s^2</span>
Answer: 71.7 KJ
Explanation:
The rotational kinetic energy of a rotating body can be written as follows:
Krot = ½ I ω2
Now, any point on the rim of the flywheel, is acted by a centripetal force, according to Newton’s 2nd Law, as follows:
Fc = m. ac
It can be showed that the centripetal acceleration, is related with the angular velocity and the radius, as follows:
ac = ω2 r
We know that this acceleration has a limit value, so , we can take this limit to obtain a maximum value for the angular velocity also.
As the flywheel is a solid disk, the rotational inertia I is just ½ m r2.
Replacing in the expression for the Krot, we have:
Krot= ½ (1/2 mr2.ac/r) = ¼ mr ac = ¼ 67.0 Kg. 1.22 m . 3,510 m/s2 = 71. 7 KJ
