Answer:
475 m , 950 m
Explanation:
Let l be the length of the side perpendicular to the barn.
1900-2l = length of the side parallel to the barn
Area A= l( 1900-2l)
A= 1900l-2l^2
now, the maximum value of l ( the equation being quadratic)
l_max= -b/2a
a= 2
b=1900
l_max= -1900/4= 475 m
then 1900-2l= 1900-2×(475) = 950 m
So, the dimensions that maximize area are
950 and 475
Now. A_max = -2( l_max)^2+1900×l_max
A_max= -2(475)^2+1900×475
A_max= 451250 m^2
or, 475×950 = 451250 m^2
Answer:
Value of 
Explanation:
We have given
In first case resistance is
and current is 1.8 A
Let the potential difference is v
So
----eqn 1
In second case resistance is
and current is 1.6 A and potential difference will be as it is a series connection
So
----eqn 2
From eqn 1 and eqn 2


From laws of motion:

Where S is the distance/displacement (as you would call it) which is unknown
v = final velocity which is 0m/s (this is because the car stops)
u = initial velocity which is 36m/s (from the data given)
t = time taken for the distance to be covered and it is 6s
Substitute the values, hence:


But this is merely the distance he travelled in the 6 seconds he was trying to stop the car.
Therefore, the distance between the car and the cows = 160-108
Distance = 52m
The phase is called 3rd quarter.
Hope this helps:)