Answer: D. The object would have an acceleration of 6 m/s/s to the right.
Explanation:
I got it right on USA test prep
The time taken for him to move the bin 6.5 m is 2.30 s.
The given parameters;
- <em>weight of the load, w = 557 N</em>
- <em>force applied , F = 410 N</em>
- <em>angle of force, = 15°</em>
- <em>coefficient of kinetic friction = 0.46</em>
- <em>distance moved, d = 6.5 m</em>
The net horizontal force on the recycling bin is calculated as follows;
where;
- <em>m is the mass of the recycling bin</em>
- <em />
<em> is the frictional force </em>
W = mg
The net horizontal force on the recycling bin is calculated as;
The time taken for him to move the bin 6.5 m is calculated as follows;
Thus, the time taken for him to move the bin 6.5 m is 2.30 s.
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When 6210 bucks is converted to kilobucks, the result obtained is 6.21 kilobucks
<h3>Conversion scale </h3>
1000 bucks = 1 Kilobuck
Using the above convesion scale, we can express 6210 bucks in kilobucks
<h3>How to express bucks in kilobucks</h3>
1000 bucks = 1 Kilobuck
Therefore,
6210 bucks = (6210 × 1) / 1000
6210 bucks = 6.21 kilobucks
Thus, 6210 bucks is equivalent to 6.21 kilobucks
Learn more about conversion:
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Displacement depends upon the path taken as it is a vector.
From your problem above we would have a total displacement of;
Defining +x direction as east and -x direction as west
6east-3west+1east+6east-7west
6-3+1+6-7=3 blocks east or + x-direction
So even though they walked a total of 17 blocks it ends up only being 3 blocks total in +xdirection that was travelled by displacement.
Any questions please ask.
Answer:
A. We have that radius r = 4.00m intensity I = 8.00 W/m^
total power = power/ Area ( 4πr2)= 8.00 w/m^2( 4π ( 4.00 m)2=1607.68 W
b) I = total power/ 4πr2= 8.00 W/m2 ( 4.00 m/ 9.5 m)2= 1.418 W/m2
c) E = total power x time= 1607 . 68 W x 1s= 1607.68 J
