1. Vpa = 180m/s. @ 0 deg.
Vag = 40m/s @ 120 deg,CCW.
<span>
Vpg = Vpa + Vag,
Vpg = (180 + 40cos120) + i40sin120,
Vpg = 160 + i34.64,
Vpg=sqrt((160)^2 + (34.64)^2)=163.7m/s.
</span>
<span>2. tanA = Y / X = 34.64 / 160 = 0.2165,
A = 12.2 deg,CCW. = 12.2deg. North of
East. </span>
3. 1 hr = 3600s. <span>d = Vt = 163.7m/s * 3600s = 589,320m.
hope this helps</span>
The driver is telling the truth, the radar gun must have been set incorrectly to record relative velocity.
The given parameters:
- <em>Speed of the driver observed by the stationary police officer, Vo = 44.7 m/s</em>
- <em>Speed of the driver, V = 26.8 m/s.</em>
- <em>Speed limit = 60 mph</em>
The speed limit of the driver in meter per second is calculated as follows;

From the speed limit, it is obvious that the driver's speed is within the limit. Thus, we can conclude that the driver is telling the truth, the radar gun must have been set incorrectly to record relative velocity.
Learn more about relative velocity here: brainly.com/question/17228388
<span>The answer is none. According to the first law of Newton, an object stays at the same speed in the same direction if there are not forces unbalancing the object. Without friction, the car would be moving forever, unless there is another force accelerating or stopping the car.</span>
work done=446.9 J . so option (c) is correct.
Explanation:
the formula for work done is given by
W= F d
F= force= mg where m= mass and g= acceleration due to gravity
F= 3.8 (9.8)=37.24 J
so W=37.24 (12)
W=446.9 J
D. adding one student to the team on the left