To solve this problem we will apply the concepts given from the circular movement of the bodies for which we have that the centripetal Force is defined as a product between the mass and the velocity squared at the rate of rotation, mathematically this is
Where,
m = Mass
v = Velocity
r = Radius
Our values are given as
Rearranging to find the velocity we have that,
Therefore the maximum speed can the mass have if the string is not to break is 29m/s
Answer:
Please refer to the attached schematic for free-body diagrams of the blocks.
The contact forces N1 = mg, N2 = 3mg.
The friction force fs = mg/2.
The tension force T = 3mg/2.
The distance y = (gt^2)/4
Explanation:
Starting with the third block which is declining a time t. Newton's Second Law implies that
There are two unknowns in one equation: T and a. Therefore, we need to find more equations to solve these unknown variables.
Let's look at free-body diagram 1:
Newton's Second Law:
That is another equation with one more unknown: fs.
Free-body diagram 2:
That is our third equation. We now have three equations with three unknowns. Combining equation 2 and 3 gives:
Plugging this equation into the first equation gives
Following this, we can find the other unknowns:
The contact forces are N1 and N2.
Finally, using the acceleration and equation of kinematics, we can find the distance the hanging weight drops in a time t.
Answer:
25.8 lb/in²
Explanation:
Gay-Lussac's law tells us that given an ideal gas of a certain mass has a constant volume, the pressure exerted on the sides of its container is directly proportional to its absolute temperature.
Apply the law of conservation of momentum for this situation. The law states that the momentum of a system is constant (in absence of external forces acting on it).
The 'system' in this case are the two skaters. There is no external force on the skaters. Suppose the skaters are initially standing still. The momentum in the system is 0. This value will need to remain constant, even after the mutual push (which is a set of forces from <em>inside</em> the system). So we know that
(total momentum before) = (total momentum after)
Indexing the masses and velocities by the first letter of the skaters' names:
From the last row, you can see that the skaters will have momentum of same magnitude but opposite direction, after the push off. That answers the first question: neither will have a greater momentum (both will have one of same magnitude).
Since Ricardo is heavier, from the above equality it follows that
In words, Paula has the greater speed, after the push-off.