Answer:
a) L=0. b) L = 262 k ^ Kg m²/s and c) L = 1020.7 k^ kg m²/s
Explanation:
It is angular momentum given by
L = r x p
Bold are vectors; where L is the angular momentum, r the position of the particle and p its linear momentum
One of the easiest ways to make this vector product is with the use of determinants
![{array}\right] \left[\begin{array}{ccc}i&j&k\\x&y&z\\px&py&pz\end{array}\right]](https://tex.z-dn.net/?f=%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5Cx%26y%26z%5C%5Cpx%26py%26pz%5Cend%7Barray%7D%5Cright%5D)
Let's apply this relationship to our case
Let's start by breaking down the speed
v₀ₓ = v₀ cosn 45
voy =v₀ sin 45
v₀ₓ = 9 cos 45
voy = 9 without 45
v₀ₓ = 6.36 m / s
voy = 6.36 m / s
a) at launch point r = 0 whereby L = 0
. b) let's find the position for maximum height, we can use kinematics, at this point the vertical speed is zero
vfy² = voy²- 2 g y
y = voy² / 2g
y = (6.36)²/2 9.8
y = 2.06 m
Let's calculate the angular momentum
L= ![\left[\begin{array}{ccc}i&j&k\\x&y&0\\px&0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5Cx%26y%260%5C%5Cpx%260%260%5Cend%7Barray%7D%5Cright%5D)
L = -px y k ^
L = - (m vox) (2.06) k ^
L = - 20 6.36 2.06 k ^
L = 262 k ^ Kg m² / s
The angular momentum is on the z axis
c) At the point of impact, at this point the height is zero and the position on the x-axis is the range
R = vo² sin 2θ / g
R = 9² sin (2 45) /9.8
R = 8.26 m
L = ![\left[\begin{array}{ccc}i&j&k\\x&0&0\\px&py&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5Cx%260%260%5C%5Cpx%26py%260%5Cend%7Barray%7D%5Cright%5D)
L = - x py k ^
L = - x m voy
L = - 8.26 20 6.36 k ^
L = 1020.7 k^ kg m² /s
Answer:
change in momentum, 
Average Force, 
Explanation:
Given:
angle of kicking from the horizon, 
velocity of the ball after being kicked, 
mass of the ball, 
time of application of force, 
We know, since body is starting from the rest
.....................(1)
Now the components:
similarly
also, impulse
.........................(2)
where F is the force applied for t time.
Then from eq. (1) & (2)
Now, the components
&
Consider a long train moving at speed v. Now consider a passenger throwing a ball inside this train, towards the back of the train, with same velocity v (but in the opposite direction of the train movement).
- A passenger inside the train will see the ball moving with speed v
- For an observer outside the train, however, the ball will appear as still. In fact, for him the ball will have a speed v (given by the movement of the train) -v (velocity of the ball but moving in the opposite direction), so the net velocity will be v+(-v)=0.
The voltage in the extension cord is 30 V.
The problem above can be solved using ohm's law
⇒ Formula:
V = IR.................. Equation 1
⇒ Where:
- V = Voltage in the extension cord
- I = Current flowing through the extension cord
- R = Resistance of the extension cord.
From the question, I think there was a slight error in the value of the current given it suppose to be 500 A, and not 5.00 A
⇒ Given:
⇒ Substitute these values into equation 1
Hence the voltage in the extension cord is 30 V
Learn more about voltage here: brainly.com/question/4429782
A force can be considered a push or pull
hope this helps :)
