Explanation:
1. draught
2. Parallax error
3. angle if displacement
4. air resistance or any form of obstruction
Given:
u = 10⁵ m/s, the entrance velocity
v = 2.5 x 10⁶ m/s, the exit velocity
s = 1.6 cm = 0.016 m, distance traveled
Let a = the acceleration.
Then
u² + 2as = v²
(10⁵ m/s)² + 2*(a m/s²)*(0.016 m) = (2.5 x 10⁶ m/s)²
0.032a = 6.25 x 10¹² - 10¹⁰ = 6.24 x 10¹²
a = 1.95 x 10¹⁴ m/s²
Answer: 1.95 x 10¹⁴ m/s²
D is the best answer. In many physics problems we treat an extended object as if it were a point with the same mass located at the center of mass.
1). The little projectile is affected by friction all the way through the block.
Friction robs some kinetic energy.
2). The block is affected by friction as it scrapes along the top of the post.
Friction robs some kinetic energy.
3). The block is also affected by friction with the air (air resistance) as it
falls to the ground. Friction robs some kinetic energy.
Explanation:
(a) Displacement of an object is the shortest path covered by it.
In this problem, a student is biking to school. She travels 0.7 km north, then realizes something has fallen out of her bag. She travels 0.3 km south to retrieve her item. She then travels 0.4 mi north to arrive at school.
0.4 miles = 0.64 km
displacement = 0.7-0.3+0.64 = 1.04 km
(b) Average velocity = total displacement/total time
t = 15 min = 0.25 hour

Hence, this is the required solution.