Answer:
Original speed of the mess kit = 4.43 m/s at 50.67° north of east.
Explanation:
Let north represent positive y axis and east represent positive x axis.
Here momentum is conserved.
Let the initial velocity be v.
Initial momentum = 4.4 x v = 4.4v
Velocity of 2.2 kg moving at 2.9 m/s, due north = 2.9 j m/s
Velocity of 2.2 kg moving at 6.8 m/s, 35° north of east = 6.9 ( cos 35i + sin35 j ) = 5.62 i + 3.96 j m/s
Final momentum = 2.2 x 2.9 j + 2.2 x (5.62 i + 3.96 j) = 12.364 i + 15.092 j kgm/s
We have
Initial momentum = Final momentum
4.4v = 12.364 i + 15.092 j
v =2.81 i + 3.43 j
Magnitude
Direction
50.67° north of east.
Original speed of the mess kit = 4.43 m/s at 50.67° north of east.
Answer:
L = 41.09 Kg m2 / s The angular momentum does not depend on the time
Explanation:
The definition of angular momentum is
L = r x p
Where blacks indicate vectors
Let's apply this definition our case. Linear momentum
p = m v
Let's replace
L = m r x v
The given function is
x = 6.00 i ^ + 4.15 t j
^
We look for speed
v = dx / dt
v = 0 + 4.15 j ^
To evaluate the angular momentum one of the best ways is to use determinants
![L = m \left[\begin{array}{ccc}i&j&k\\6&4.15t&0\\0&4.15&0\end{array}\right]](https://tex.z-dn.net/?f=L%20%3D%20m%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C6%264.15t%260%5C%5C0%264.15%260%5Cend%7Barray%7D%5Cright%5D)
L = m 6 4.15 k ^
The other products give zero
Let's calculate
L = 1.65 6 4.15 k ^
L = 41.09 Kg m2 / s
The angular momentum does not depend on the time
There are only two things you can do to increase the work output of a machine:
1). Increase the work INput to the machine.
2). Make the machine more efficient ... do things like lubricating it better to eliminate some internal friction.
Answer:
this is a law because it is a constant fact of nature
Explanation:
Answer:
b. negative
Explanation:
neutrons have a negative charge and protons have a proton has a positive charge
