Answer:
The total momentum of the system before the collision is 5.334 kg-m/s towards left.
Explanation:
Given that,
Mass of the block A, 
Speed of block A, 
Mass of the block B, 
Mass of block B, 
Let p is the total momentum of the system before the collision. It is given by :
So, the total momentum of the system before the collision is 5.334 kg-m/s towards left. Hence, this is the required solution.
The airplane's speed relative to the ground is
√ (100² + 25²)
= √ (10,000 + 625)
= √ 10,625
= 103.08 km/hr .
The angle of its velocity north of west is
the angle whose tangent is (25/100)
arctan(25/100) = 14° north of west .
(bearing = 284°)
Answer:
total time = 65 seconds
total distance = 1554 meters
Explanation:
kinematic equation:
final velocity = initial velocity + acceleration multiplied by time
v_1 = v_0 + at
28 m/s = 0 m/s + 2 m/s^2 (t)
t = 14 seconds
a) total time = 14 + 46 + 5 = 65 seconds
b) must solve for total distance and divide it by time.
d_1 = v_0t + 1/2 a * t^2
d_1 = 0 + 0.5(2) * 14^2
d_1 = 196 meters
d2 = vt
d2 = 28 *46
d2 = 1288 meters
v_1 = v_o + at
0 = 28 + a(5)
- 28/5 = a
a = - 5.6 m/s^2
d_3 = v_0t + 1/2 a * t^2
d_3 = 28 (5) - 0.5(5.6)*5^2
d_3 = 70 meters
total distance = d1 + d2 + d3 = 196 + 1288 + 70 = 1554 meters
Your answer is B
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