Answer:
I can't understand your question sorry
Solution:
According to the equations for 1-D kinematics. The only change to them is that instead one equation that describes general motion.
So we will have to use the equations twice: once for motion in the x direction and another time for the y direction.
v_f=v_o + at ……..(a)
[where v_f and v_o are final velocity and initial velocity, respectively]
Now ,
Initially, there was y velocity, however gravity began to act on the football, causing it to accelerate.
Applying this value in equation (a)
v_yf = at = -9.81 m/s^s * 1.75 = -17.165 m/s in the y direction
For calculating the magnitude of the equation we have to square root the given value
(16.6i - 17.165y)
\\
\left | V \right |=sqrt{16.6^{2}+17.165^{2}}\\ =
\sqrt{275.56+294.637225}\\=
\sqrt{570.197225}\\=
23.87[/tex]
On a similar problem wherein instead of 480 g, a 650 gram of bar is used:
Angular momentum L = Iω, where
<span>I = the moment of inertia about the axis of rotation, which for a long thin uniform rod rotating about its center as depicted in the diagram would be 1/12mℓ², where m is the mass of the rod and ℓ is its length. The mass of this particular rod is not given but the length of 2 meters is. The moment of inertia is therefore </span>
<span>I = 1/12m*2² = 1/3m kg*m² </span>
<span>The angular momentum ω = 2πf, where f is the frequency of rotation. If the angular momentum is to be in SI units, this frequency must be in revolutions per second. 120 rpm is 2 rev/s, so </span>
<span>ω = 2π * 2 rev/s = 4π s^(-1) </span>
<span>The angular momentum would therefore be </span>
<span>L = Iω </span>
<span>= 1/3m * 4π </span>
<span>= 4/3πm kg*m²/s, where m is the rod's mass in kg. </span>
<span>The direction of the angular momentum vector - pseudovector, actually - would be straight out of the diagram toward the viewer. </span>
<span>Edit: 650 g = 0.650 kg, so </span>
<span>L = 4/3π(0.650) kg*m²/s </span>
<span>≈ 2.72 kg*m²/s</span>
Answer:
a) 
Now we can replace the velocity for t=1.75 s
For t = 3.0 s we have:
b) 
And we can find the positions for the two times required like this:
And now we can replace and we got:
Explanation:
The particle position is given by:
Part a
In order to find the velocity we need to take the first derivate for the position function like this:
Now we can replace the velocity for t=1.75 s
For t = 3.0 s we have:
Part b
For this case we can find the average velocity with the following formula:
And we can find the positions for the two times required like this:
And now we can replace and we got:
