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:
Walking’ - If a handball player takes more than three steps without dribbling (bouncing the ball) or holds the ball for more than 3 seconds without bouncing it, shooting or passing, then that is deemed ‘walking' and possession is lost.
'Double dribble’ - Handball players cannot receive the ball and bounce it, then hold the ball, and bounce it again. This is termed ‘double dribble’ and is against the rules.
Askmeanything2♡
Power delivered = (energy delivered) / (time to deliver the energy)
Power delivered = (4,000 J) / (0.5 sec)
Power delivered = 8,000 watts
I'm a little surprised to learn that Electro draws his power from the mains. This is VERY good news for Spiderman ! It means that Spiderman can always avoid tangling with Electro ... all he has to do is stay farther away from Electro than the length of Electro's extension cord.
But OK. Let's assume that Electro draws it all from the mains. Then inevitably, there must be some loss in Electro's conversion process, between the outlet and his fingertips (or wherever he shoots his bolts from).
The efficiency of Electro's internal process is
<em>(power he shoots out) / (power he draws from the mains) </em>.
So, if he delivers energy toward his target at the rate of 8,000 watts, he must draw power from the mains at the rate of
<em>(8,000 watts) / (his internal efficiency) . </em>
a) The kinetic energy (KE) of an object is expressed as the product of half of the mass (m) of the object and the square of its velocity (v²):

It is given:
v = 8.5 m/s
m = 91 kg
So:

b) We can calculate height by using the formula for potential energy (PE):
PE = m*g*h
In this case, h is eight, and PE is the same as KE:
PE = KE = 3,287.4 J
m = 91 kg
g = 9.81 m/s² - gravitational acceleration
h = ? - height
Now, let's replace those:
3,287.4= 91 * 9.81 * h
⇒ h = 3,287.4/(91*9.81) = 3,287.4/892.7 = 3.7 m
Answer:
<h3>b.fission. </h3>
Explanation:
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