Answer:
Rectangular path
Solution:
As per the question:
Length, a = 4 km
Height, h = 2 km
In order to minimize the cost let us denote the side of the square bottom be 'a'
Thus the area of the bottom of the square, A = 
Let the height of the bin be 'h'
Therefore the total area, 
The cost is:
C = 2sh
Volume of the box, V =
(1)
Total cost,
(2)
From eqn (1):

Using the above value in eqn (1):


Differentiating the above eqn w.r.t 'a':

For the required solution equating the above eqn to zero:


a = 4
Also

The path in order to minimize the cost must be a rectangle.
Answer:
Explanation:
Givens
vi = 10 m/s
a = 1.5 m/s^2
d = 600 m
vf = ?
Formula
vf^2 = vi^2 + 2*a*d
Solution
vf^2 = 10^2 + 2*1.5 * 600
vf^2 = 100 + 1800
vf^2 = 1900
sqrt(vf^2) = sqrt(1900)
vf = 43.59 m/s
Answer:
The magnitude of F₁ is 3.7 times of F₂
Explanation:
Given that,
Time = 10 sec
Speed = 3.0 km/h
Speed of second tugboat = 11 km/h
We need to calculate the speed


The force F₁is constant acceleration is also a constant.

We need to calculate the acceleration
Using formula of acceleration



Similarly,

For total force,


The speed of second tugboat is


We need to calculate total acceleration



We need to calculate the acceleration a₂



We need to calculate the factor of F₁ and F₂
Dividing force F₁ by F₂



Hence, The magnitude of F₁ is 3.7 times of F₂
Answer:

Explanation:
initially the merry go round is at rest
after 6.73 s the merry go round will accelerates to 20 rpm
so final angular speed is given as



so final tangential speed is given as


now average acceleration of the girl is given as



Explanation:
velocity is distance divided by time.
so
the average speed of the ball is 10m/20s
= 0.5 m/s