Answer:
Explanation:36.05 km
Given
First car travels
South
then turns and travels
east
Suppose south as negative y axis and east as positive x axis
So, 

Displacement is the shortest between initial and final point
Dispalcement
Displacement
Displacement
Magnitude 
Magnitude
Answer:
20.25 m
Explanation:
- <u>Centripetal acceleration </u>is given by; the square of the velocity, divided by the radius of the circular path.
That is;
<em><u>ac = v²/r</u></em>
<em> </em><em><u> Where; ac = acceleration, centripetal, m/s², v is the velocity, m/s and r is the radius, m</u></em>
Therefore;
r = v²/ac
= 27²/36
= 20.25 m
Hence the radius is 20.25 meters
Answer:
Explanation:
This is a simple gravitational force problem using the equation:
where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.
Filling in:
I'm going to do the math on the top and then on the bottom and divide at the end.
and now when I divide I will express my answer to the correct number of sig dig's:
6.45 × 10¹⁶ N
Use the distance swan and the time elapsed in that interval.
Average velocity = distance / time
Average velocity = [4.0 m + 3.0m] / 3.2 s = 2.1875 m/s
Answer:
The coefficient of kinetic friction between the puck and the ice is 0.11
Explanation:
Given;
initial speed, u = 9.3 m/s
sliding distance, S = 42 m
From equation of motion we determine the acceleration;
v² = u² + 2as
0 = (9.3)² + (2x42)a
- 84a = 86.49
a = -86.49/84
|a| = 1.0296
= ma
where;
Fk is the frictional force
μk is the coefficient of kinetic friction
N is the normal reaction = mg
μkmg = ma
μkg = a
μk = a/g
where;
g is the gravitational constant = 9.8 m/s²
μk = a/g
μk = 1.0296/9.8
μk = 0.11
Therefore, the coefficient of kinetic friction between the puck and the ice is 0.11