An elevator starts from rest with a constant. upward acceleration andmoves 1m in the first. 1.8 s. a passenger in the elevator is holding a. 8.7 kg bundle at the end of a vertical cord.. what is the tension in the cord as the ele-. vator accelerates? the acceleration of gravity. is 9.8 m/s2 .. answer in units of n

Newton's second law tells you: 

Sum of forces on an object = ma 

Here, the forces acting on the bundle are the tension in the string and the force of gravity, these two must combine to yield the acceleration of the bundle. 

So we have: 

T-mg = ma 

or T=m(g+a) 

We know m=8.7kg, we need to find a from the information 

starting from rest, an accelerating object covers distance according to: 

dist = 1/2 at^2 

to cover 1m in 1.8s, we have: 

a=2d/t^2 = 2x1/1.8^2 = 0.62 m/s/s 

Thus, the tension in the string is: 

T = m(g+a)
=  8.7kg(9.8m/s/s+0.62m/s/s) 

T =  90.654 N

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The speed was not exceeded so it was slow

Answerthe force between the charges will be  f/9 (one-ninth of force f)explanationcoulomb's law relates the magnitudes  two point charges q₁ and q₂, the distance, r, between the charges, and the electrostatic force, f, between the two charges:     k represents coulomb's constant, k = 8.99×10⁹ n·m²/c².note that if we have r = 3r, then