I notice that even though we're working with frames of reference
here, you never said which frame the '5 km/hr' is measured in.
In fact ! You didn't even say which frame the '12 km/hr' of his
bike is measured in.
So there are several different ways this could go. I'll do it the way
I THINK you meant it, but that doesn't guarantee anything.
-- Simon is riding his bike at 12 km/hr relative to the sidewalk,
away from Keesha.
-- He throws a ball at Keesha, at 5 km/hr relative to his own face.
-- Keesha sees the ball approaching her at (12 - 5) = 7 km/hr
relative to the ground and to her.
Mostly gravity voloume and sometimes what it is made of
Answer:
ω = 1.83 rad/s clockwise
Explanation:
We are given:
I1 = 3.0kg.m2
ω1 = -5.4rad/s (clockwise being negative)
I2 = 1.3kg.m2
ω2 = 6.4rad/s (counterclockwise being positive)
By conservation of the momentum:
I1 * ω1 + I2 * ω2 = (I1 + I2) * ω
Solving for ω:

Since it is negative, the direction is clockwise.
C. have similar properties
(this is because they have the same number of electrons in the outer orbital)
Answer:
Displacement: 6.71 m, Direction: 63.4 degrees north of east
Explanation:
In the attached image we can aprecciate each one of the movements of the parade. Let's say that the parade started from the origin (point (0,0)) then it moves to the east 4 blocks it means now the parade is located at point (4,0).
Then the parade went to the south three blocks, so it moves to the coordinate (4,-3). After this the parade went to the west one block so the new coordinate point is (3, -3).
And finally the movement of the 0 parade was 9 blocks to the north. It means the final point is now (0,9) - (3,-3) = (3,6)
And the displacement will be defined by the folliwing vector operation:

We know that the magnitude of the displacement vector is defined by the phytagoras theorem

And the angle will be defined by:
tan(beta)=3/6
beta = tan^-1(6/3)
beta = 63.43°