<span>As seen by Barbara, Neil is traveling at a velocity of 6.1 m/s at and angle of 76.7 degrees north from due west.
Let's assume that both Barbara and Neil start out at coordinate (0,0) and skate for exactly 1 second. Where do they end up?
Barbara is going due south at 5.9 m/s, so she's at (0,-5.9)
Neil is going due west at 1.4 m/s, so he's at (-1.4,0)
Now to see Neil's relative motion to Barbara, compute a translation that will place Barbara back at (0,0) and apply that same translation to Neil. Adding (0,5.9) to their coordinates will do this.
So the translated coordinates for Neil is now (-1.4, 5.9) and Barbara is at (0,0).
The magnitude of Neil's velocity as seen by Barbara is
sqrt((-1.4)^2 + 5.9^2) = sqrt(1.96 + 34.81) = sqrt(36.77) = 6.1 m/s
The angle of his vector relative to due west will be
atan(5.9/1.4) = atan(4.214285714) = 76.7 degrees
So as seen by Barbara, Neil is traveling at a velocity of 6.1 m/s at and angle of 76.7 degrees north from due west.</span>
Answer:
ω = 630.2663 = 630[rad/s]
Explanation:
Solution:
- We can tackle this question by simple direct proportion relation between angular speed for the disk to rotate a cycle that constitutes 20 holes. We will use direct relation with number of holes per cycle to compute the revolution per seconds i.e frequency of speed f.
1rev(20 hole) -> 20(cycle)/rev
2006.2(cycle) -> f ?
f = 2006.2/20 = 100.31rev at second
- The relation between angular frequency and angular speed is given by:
ω = 2πf
ω = 2*3.14*100.31
ω = 630.2663 = 630[rad/s]
Since toy is moving at constant speed that means that force that child is applying on toy is equal to force of friction.
Rate of speed that toy is moving is irelevant.
childs force is:
Fc = 2N
Fc = Ff (Ff -friction force)
Ff = a*Q
where Q is weight of the toy and a is friction
if we express a we get
a = F/Q = 2/8 = 0.25
The statements that apply in this case are:
They show the elements that make up a compound.
They show the types of atoms that make up a molecule.
They show the number of each type of atom in a molecule.
Answer:
We need 4 times more force to keep the car in circular motion if the velocity gets double.
Explanation:
Lets take the mass of the car = m
The radius of the arc = r
Given that speed of the car gets double ,v' = 2 v
Then the force on the car = F'
( radius of the arc is constant)
We know that
Therefore F' = 4 F
So we can say that we need 4 times more force to keep the car in circular motion if the velocity gets double.
