An object of mass 400 kg islocated at < 86, -25, 0> m is moving at avelocity of < 6, -9,0> m/s. Two forcesF₁ and F₂ are applied on the object. Thenet force of < 90, -75,0> N is applied for 13 s.2If F₂ = < 80, -90,0 > N, calculate1)F₁ = <21eIn the formula sheet provided,wuch equation can be used to calculate thethe impulse applied to the object? →3)The impulse applied to theobject during this 13 s interval is:Ap =<4)In the formula sheet provided,which equation can be used to calculate thefinal position of the object?5)the end of this 13 s interval is:The location of the particle at=<

Given,mass , m = 400 kg Initial position, r0 = <86,-25,0>mInitial velocity,v0 =<6,-9,0>…

An object of mass 400 kg is located at < 86, -25, 0> m is moving at a velocity of < 6, -9,0> m/s. Two forces F₁ and F₂ are applied on the object. The net force of < 90, -75,0> N is applied for 13 s.

An object of mass 400 kg is located at < 86, -25, 0> m is moving at a velocity of < 6, -9,0> m/s. Two forces F₁ and F₂ are applied on the object. The net force of < 90, -75,0> N is applied for 13 s.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Concept explainers

Newton’s Third Law of Motion

The horse pulls the cart and the cart pulls the horse in response as per Newton third law. So the big question is, “Don’t the two forces cancel each other? How does the cart accelerate?”

Question

An object of mass 400 kg is
located at < 86, -25, 0> m is moving at a
velocity of < 6, -9,0> m/s. Two forces
F₁ and F₂ are applied on the object. The
net force of < 90, -75,0> N is applied for 13 s.
2
If F₂ = < 80, -90,0 > N, calculate
1)
F₁ = <
21e
In the formula sheet provided,
wuch equation can be used to calculate the
the impulse applied to the object? →
3)
The impulse applied to the
object during this 13 s interval is:
Ap =<
4)
In the formula sheet provided,
which equation can be used to calculate the
final position of the object?
5)
the end of this 13 s interval is:
The location of the particle at
=<

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