There are mistakes in the question.The correct question is here
A 2.0 kg, 20-cm-diameter turntable rotates at 100 rpm on frictionless bearings. Two 500 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diameter, and stick. What is the turntable’s angular velocity, in rpm, just after this event?
Answer:
w=50 rpm
Explanation:
Given data
The mass turntable M=2kg
Diameter of the turntable d=20 cm=0.2 m
Angular velocity ω=100 rpm= 100×(2π/60) =10.47 rad/s
Two blocks Mass m=500 g=0.5 kg
To find
Turntable angular velocity
Solution
We can find the angular velocity of the turntable as follow
Lets consider turntable to be disk shape and the blocks to be small as compared to turntable

where I is moment of inertia

Answer:
1.574 [m/s²].
Explanation:
1) acceleration is: a=(V-V₀)/t, where V - final velocity [68km/h=18.(8)m/s]; V₀ - initial velocity [0]; t - elapsed time [12sec.];
2) according to the formula above:
a=18.(8)/12≈1.574 [m/s²].
Answer:
k = 1,250 N/m
Explanation:
Use the formula F=kx, with F=5N and x=0.04m
Then the spring constant (k) is 5/0.04
Answer:
See explanation
Explanation:
In this exercise, we need to use the law of conservation of angular momentum which is:
I1*W1 + I2*W2 = (I1 + I2)*W2'
Where:
I: moment of innertia.
W: angular velocity
Now let's call 1 the runner and 2, the turntable. the system would be W2'.
The angular speed of the runner, we can calculate that with the following expression:
W = V/r
so:
W1 = 2.5 / 3.6 = 0.694 rad/s
The innertia is calculated with the expression:
I = m*r²
I = 60 * 3.6 = 216 kg.m²
I2 and W2 are provided in the exercise, so, replacing all the data in the conservation of angular momentum, let's solve for W2'
(216*0.694) + (-0.190*81) = (81 + 216)W2'
134.514 = 297W2'
W2' = 134.514 / 297
W2' = 0.453 rad/s