Answer:
K_b = 78 J
Explanation:
For this exercise we can use the conservation of energy relations
starting point. Lowest of the trajectory
Em₀ = K = ½ mv²
final point. When it is at tea = 50º
Em_f = K + U
Em_f = ½ m v_b² + m g h
where h is the height from the lowest point
h = L - L cos 50
Em_f = ½ m v_b² + mg L (1 - cos50)
energy be conserve
Em₀ = Em_f
½ mv² = ½ m v_b² + mg L (1 - cos50)
K_b = ½ m v_b² + mg L (1 - cos50)
let's calculate
K_b = ½ 2.0 6.0² + 2.0 9.8 6.0 (1 - cos50)
K_b = 36 +42.0
K_b = 78 J
<span><span>Your friend is bragging about his motorcycle. He claims that it can go from a stopped position to 50 miles per hour in three seconds. He is describing the motorcycle's
</span>Answer: </span>acceleration
Hope This Helps! :3
Answer:
-142m
Explanation:
To find the initial position of the car, we can add the displacement.
-245 + 103 = -142
Best of Luck!
Https://courses.physics.illinois.edu/phys102/sp2013/lectures/lecture2.pdf
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