Answered

Westonci.ca is the premier destination for reliable answers to your questions, brought to you by a community of experts. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.

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.

Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.