Hello!
This is an example of an inelastic collision, where the two objects "stick" to each other after their collision. (The Goalkeeper CATCHES the puck).
We can write out the conservation of momentum formula:
m1vi + m2vi = m1vf + m2vf
Let:
m1 = mass of puck
m2 = mass of the goalkeeper
We know that the initial velocity of the goalkeeper is 0, so:
m1vi + m2(0) = m1vf + m2vf
m1vi = m1vf + m2vf
The final velocities will be the same, so:
m1vi = (m1 + m2)vf
Plug in the given values:
(0.16)(40)/ (0.16 + 120) = vf ≈ 0.0533 m/s
Using the equation for momentum:
p = mv
The object with the LARGER mass will have the greater momentum. Thus, the Goalkeeper has the largest momentum as p = mv; a greater mass correlates to a greater momentum since the velocity is the same between the two objects. The puck would have a momentum of p = (.16)(0.0533) = 0.008528 kgm/s, whereas the goalkeeper would have a momentum of
p = (120)(0.0533) = 6.396 kgm/s.
Answer:
Around 44.01g.
Explanation:
One mole of carbon dioxide molecules has a mass of 44.01g, while one mole of sodium sulfide formula units has a mass of 78.04g.
90.9 is the velocity of an airplane
To solve this problem it is necessary to use the conservation equations of both kinetic, rotational and potential energy.
By definition we know that
Where,
KE =Kinetic Energy
KR = Rotational Kinetic Energy
PE = Potential Energy
In this way
Where,
m = mass
v= Velocity
I = Moment of Inertia
Angular velocity
g = Gravity
h = Height
We know as well that 
for velocity (v) and Radius (r)
Therefore replacing we have
![[tex]h= \frac{1}{2} \frac{v^2}{g} +\frac{1}{2} \frac{I}{mg}(\frac{v}{r})^2](https://tex.z-dn.net/?f=%5Btex%5Dh%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7Bv%5E2%7D%7Bg%7D%20%2B%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7BI%7D%7Bmg%7D%28%5Cfrac%7Bv%7D%7Br%7D%29%5E2)
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Therefore the height must be 0.3915 for the yo-yo fall has a linear speed of 0.75m/s
Answer:
9.80665 is the value in meters per second squared
