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 ?
<h2>
Answer:</h2>
<em><u>Velocity of throwing arrow = 43.13 m/s.</u></em>
<h2>
Explanation:</h2>
In the question,
Let us say the height from which the arrow was shot = h
Distance traveled by the arrow in horizontal = 61 m
Angle made by the arrow with the ground = 2°
So,
From the <u>equations of the motion</u>,
Now,
Also,
Finally, the angle made is 2 degrees with the horizontal.
So,
Final horizontal velocity = v.cos20°
Final vertical velocity = v.sin20°
Now,
u = v.cos20° (No acceleration in horizontal)
Also,
So,
We can say that,
<em><u>Therefore, the velocity with which the arrow was shot by the archer is 43.13 m/s.</u></em>
Answer:
Explanation:
Given that
Radius =r
Electric filed =E
Q=Charge on the ring
The electric filed at distance x given as
For maximum condition
For maximum condition
At the electric field will be maximum.
Let the Blaise runs for time "t" to complete the race
so the total distance he moved is given by
Now Issac runs for time t = "t - 2*60"
because it took rest for 2 minutes
now it is given that Blaise wins by 10 m distance
now the distance moved by Blaise is given by
