I am pretty sure it is B....
We have to add two vectors.
Vector #1: 0.15 m/s north
Vector #2: 1.50 m/s east
Their sum:
Magnitude: √(0.15² + 1.50²)
Magnitude = √(0.0225+2.25)
Magnitude = √2.2725
Magnitude = <em>1.5075 m/s</em>
Direction = arctan(0.15/1.50) north of east
Direction = <em>5.71° north of east</em>
To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.
From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

Where,
Angular velocity
v = Lineal Velocity
R = Radius
At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

Where
Angular acceleration
Angular velocity
t = Time
Our values are




Replacing at the previous equation we have that the angular velocity is



Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s
At the same time the angular acceleration would be



Therefore the angular acceleration of a point on the outer edge of the tires is 
you can check attachment for answer.
kind regards
Answer:
E = 10⁵ J
Explanation:
given,
Power, P = 100 TW
= 100 x 10¹² W
time, t = 1 ns
= 1 x 10⁻⁹ s
The energy of a single pulse is:-
Energy = Power x time
E = P t
E = 100 x 10¹² x 1 x 10⁻⁹
E = 10⁵ J
The energy contained in a single pulse is equal to 10⁵ J