<h2><em>So there is two truths given. After an amount of time Ttotal (lets call it ‘t’):
</em></h2><h2><em>
</em></h2><h2><em>The car’s speed is 25m/s
</em></h2><h2><em>The distance travelled is 75m
</em></h2><h2><em>Then we have the formulas for speed and distance:
</em></h2><h2><em>
</em></h2><h2><em>v = a x t -> 25 = a x t
</em></h2><h2><em>s = 0.5 x a x t^2 -> 75 = 0.5 x a x t^2
</em></h2><h2><em>Now, we know that both acceleration and time equal for both truths. So we can say:
</em></h2><h2><em>
</em></h2><h2><em>t = 25 / a
</em></h2><h2><em>t^2 = 75 / (0.5 x a) = 150 / a
</em></h2><h2><em>Since we don’t want to use square root at 2) we go squared for 1):
</em></h2><h2><em>
</em></h2><h2><em>t^2 = (25 / a) ^2 = 625 / a^2
</em></h2><h2><em>t^2 = 150 / a
</em></h2><h2><em>Since t has the same value for both truths we can say:
</em></h2><h2><em>
</em></h2><h2><em>625 / a^2 = 150 / a
</em></h2><h2><em>
</em></h2><h2><em>Thus multiply both sides with a^2:
</em></h2><h2><em>
</em></h2><h2><em>625 = 150 x a, so a = 625 / 150 = 4.17
</em></h2><h2><em>
</em></h2><h2><em>We can now calculate t as well t = 25 * 150 / 625 = 6</em></h2>
Hey there!
Your answer: Spilling breaker
Spilling breaker usually occurs when a beach or ocean is flat, and as the waves of the wind continues to happen, slowly the region would eventually become a slope.
It's almost like play-dough. Let's say that we set a perfect flat surface of play-dough on the table. As we continue slide our hands one direction, doesn't the play dough have more on one side than the other? It eventually contains a slope if you add enough from the first place.
Your answer: Spilling breaker
To solve this problem it is necessary to apply the concepts related to mutual inductance in a solenoid.
This definition is described in the following equation as,
Where,
permeability of free space
Number of turns in solenoid 1
Number of turns in solenoid 2
Cross sectional area of solenoid
l = Length of the solenoid
Part A )
Our values are given as,
Substituting,
PART B) Considering that many of the variables remain unchanged in the second solenoid, such as the increase in the radius or magnetic field, we can conclude that mutual inducantia will appear the same.
Work done by a given force is given by
here on sled two forces will do work
1. Applied force by Max
2. Frictional force due to ground
Now by force diagram of sled we can see the angle of force and displacement
work done by Max = 
Now similarly work done by frictional force
Now total work done on sled
