I think it’s the first one
I'll bite:
-- Since the sled's mass is 'm', its weight is 'mg'.
-- Since the coefficient of kinetic friction is μk, the force acting opposite to the direction it's sliding is (μk) times (mg) .
-- If the pulling force is constant 'F', then the horizontal forces on the sled
are 'F' forward and (μk · mg) backwards.
-- The net force on the sled is (F - μk·mg).
(I regret the visual appearance that's beginning to emerge,
but let's forge onward.)
-- The sled's horizontal acceleration is (net force) / (mass) = (F - μk·mg) / m.
This could be simplified, but let's not just yet.
-- Starting from rest, the sled moves a distance 's' during time 't'.
We know that s = 1/2 a t² , and we know what 'a' is. So we can write
s = (1/2 t²) (F - μk·mg) / m .
Now we have the distance, and the constant force.
The total work is (Force x distance), and the power is (Work / time).
Let's put it together and see how ugly it becomes. Maybe THEN
it can be simplified.
Work = (Force x distance) = F x (1/2 t²) (F - μk·mg) / m
Power = (Work / time) = <em>F (t/2) (F - μk·mg) / m </em>
Unless I can come up with something a lot simpler, that's the answer.
To simplify and beautify, make the partial fractions out of the
2nd parentheses:
<em> F (t/2) (F/m - μk·m)</em>
I think that's about as far as you can go. I tried some other presentations,
and didn't find anything that's much simpler.
Five points,ehhh ?
Answer:
<h2>
B. Switch</h2>
Explanation:
<u>A device designed to open or close a circuit under controlled conditions is called a switch. The terms “open” and “closed” refer to switches as well as entire circuits. An open switch is one without continuity: current cannot flow through it.</u>
<h2><u>
Hope this helps! Please consider marking brainliest!! </u></h2>
F=ma
a=(v2-v1)/(t2-t1)
a=(6-0)/(12-0)
a=6/12
a= .5 m/s^2
f=2300kg*.5m/s^2
f=1150N
f=1200N if using correct sig figs
A force over distance is work the unite is joules
