Answer:
a) 
b) 
c) 
Explanation:
<em><u>The knowable variables are </u></em>




Since the three traffic signs are <u>equally spaced</u>, the <u>distance between each sign is
</u>
a) 
b) 
Since we know the velocity in two points and the time the car takes to pass the traffic signs
c) 
Answer:
I am not a driver, but I think it's C.
Explanation:
Answer:
Final velocity of the block = 2.40 m/s east.
Explanation:
Here momentum is conserved.
Initial momentum = Final momentum
Mass of bullet = 0.0140 kg
Consider east as positive.
Initial velocity of bullet = 205 m/s
Mass of Block = 1.8 kg
Initial velocity of block = 0 m/s
Initial momentum = 0.014 x 205 + 1.8 x 0 = 2.87 kg m/s
Final velocity of bullet = -103 m/s
We need to find final velocity of the block( u )
Final momentum = 0.014 x -103+ 1.8 x u = -1.442 + 1.8 u
We have
2.87 = -1.442 + 1.8 u
u = 2.40 m/s
Final velocity of the block = 2.40 m/s east.
Answer:
24.3 degrees
Explanation:
A car traveling in circular motion at linear speed v = 12.8 m/s around a circle of radius r = 37 m is subjected to a centripetal acceleration:

Let α be the banked angle, as α > 0, the outward centripetal acceleration vector is split into 2 components, 1 parallel and the other perpendicular to the road. The one that is parallel has a magnitude of 4.43cosα and is the one that would make the car slip.
Similarly, gravitational acceleration g is split into 2 component, one parallel and the other perpendicular to the road surface. The one that is parallel has a magnitude of gsinα and is the one that keeps the car from slipping outward.
So 


