Answer:
The mass of the sand that will fall on the disk to decrease the is 0.3375 kg
Explanation:
Moment before = Moment after
where;
I is moment of inertia = Mr² = 0.3 x (0.3)² = 0.027 kg.m²
substitute this in the above equation;
![m = \frac{ 0.027[3(2 \pi) - 2(2 \pi)]} {0.2^2 * 6\pi } = \frac{ 0.027[6 \pi - 4\pi]} {0.2^2 * 4\pi }\\\\m = 0.3375kg](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7B%200.027%5B3%282%20%5Cpi%29%20%20-%202%282%20%5Cpi%29%5D%7D%20%7B0.2%5E2%20%2A%206%5Cpi%20%7D%20%3D%20%5Cfrac%7B%200.027%5B6%20%5Cpi%20%20-%204%5Cpi%5D%7D%20%7B0.2%5E2%20%2A%204%5Cpi%20%7D%5C%5C%5C%5Cm%20%3D%200.3375kg)
Therefore, the mass of the sand that will fall on the disk to decrease the is 0.3375 kg
= 454.55 g/cm3
I'm not too sure since the graduated cylinder was missing and I really don't know how to do it then. But give this a shot. Are you sure it wasn't a graduated cylinder, because I have no idea what that means
Answer:
44.3 m/s
Explanation:
a) Draw a free body diagram of the mass M. There are three forces:
Weight force mg pulling down,
Normal force N pushing perpendicular to the ramp,
and tension force T pulling parallel up the ramp.
Sum of forces in the parallel direction:
∑F = ma
T − Mg sin 30° = 0
T = Mg sin 30°
T = Mg / 2
Draw a free body diagram of the hanging mass m. There are two forces:
Weight force mg pulling down,
and tension force T pulling up.
Sum of forces in the vertical direction:
∑F = ma
T − mg = 0
T = mg
Substitute:
mg = Mg / 2
m = M / 2
M = 2m
b) Velocity of a standing wave in a string is:
v = √(T / μ)
T = mg, and m = 5 kg, so T = (5 kg) (9.8 m/s²) = 49 N. Therefore:
v = √(49 N / 0.025 kg/m)
v = 44.3 m/s
