Wow! 2.95m/s is a mighty fast pace for a backpacker. Must have one of those Star Wars anti-gravity packs. Also, I would be curious as to why she passed her destination and then walked back.
<span>Anyway, it goes like this: </span>
<span>Say the time walking east is 't', and the total time is 'T'. </span>
<span>Then 5340 m + .511 t = 1.43 T </span>
<span>(This assumes that velocity is positive in both directions) </span>
<span>Two unknowns in one equation. But you also know that the time spent walking west is </span>
<span>5340m/ 2.95m/s = 1810 sec. </span>
<span>and the total time T = 1810 +t </span>
<span>Substitute this into the first equation, and you can solve for t = 3092 sec. </span>
<span>Then T = 4902 sec. and distance walked east is .511t = 1580m.</span>
From the definition of average velocity,
,
and the fact that constant acceleration means

we can solve for the time
:

Answer:
<h2>1139.5 J</h2>
Explanation:
The work done by an object can be found by using the formula
workdone = force × distance
From the question we have
work done = 265 × 4.3 = 1139.5
We have the final answer as
<h3>1139.5 J</h3>
Hope this helps you
The net electric force acting on a positive test charge at the origin is determined as ¹/₉(kq₁q₂).
<h3>
Net electric force on the charges</h3>
The net electric force on the charges is calculated as follows;
F = kq₁q₂/r²
where;
- k is coulomb's constant
- q₁ and q₂ are the charges
- r is the distance between the charges
<h3>Distance between the charges</h3>


Thus, the net electric force acting on a positive test charge at the origin is determined as ¹/₉(kq₁q₂).
Learn more about electric force here: brainly.com/question/17692887
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Answer:
A. 91 meters north
Explanation:
Take +y to be north.
Given:
v₀ = 13 m/s
a = 0 m/s²
t = 7 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (13 m/s) (7 s) + ½ (0 m/s²) (7 s)²
Δy = 91 m
The displacement is 91 m north.