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A 375 N unbalanced force was applied to an 45.0 kg object at rest. How far would this object travel after 7.00 seconds? (5 points)

Sagot :

MathPhys

Answer:

204 m

Explanation:

From Newton's second law of motion, the net force (ΣF) on an object is equal to the mass (m) times the acceleration (a). For an object moving at constant acceleration, we can use kinematic equations also known as SUVAT equations to model the motion. For this problem, we will use the equation:

s = ut + ½ at²

where

  • s is the displacement (distance and direction)
  • u is the initial velocity
  • a is the acceleration
  • t is the time

First, use Newton's second law of motion to find the acceleration.

ΣF = ma

375 N = (45.0 kg) a

a = 8.33 m/s²

Next, use the kinematic equation to find the displacement.

s = ut + ½ at²

s = (0 m/s) (7.00 s) + ½ (8.33 m/s²) (7.00 s)²

s = 204 m

Rounded to three significant figures, the displacement of the object is 204 meters.

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