The point beneath surface where rock breaks and an arthquake is produced is known as the

An offshore oil well is 2 kilometers off the coast. The refinery is 4 kilometers down the coast. Laying pipe in the ocean is twi

shusha [124]

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 = $a^{2}$

Let the height of the bin be 'h'

Therefore the total area, $A_{t} = 4ah$

The cost is:

C = 2sh

Volume of the box, V = $a^{2}h = 4^{2}\times 2 = 128$ (1)

Total cost, $C_{t} = 2a^{2} + 2ah$ (2)

From eqn (1):

$h = \frac{128}{a^{2}}$

Using the above value in eqn (1):

$C(a) = 2a^{2} + 2a\frac{128}{a^{2}} = 2a^{2} + \frac{256}{a}$

$C(a) = 2a^{2} + \frac{256}{a}$

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

$C'(a) = 4a - \frac{256}{a^{2}} = \frac{4a^{3} - 256}{a^{2}}$

For the required solution equating the above eqn to zero:

$\frac{4a^{3} - 256}{a^{2}} = 0$

a = 4

Also

$h = \frac{128}{4^{2}} = 8$

The path in order to minimize the cost must be a rectangle.

8 0

3 years ago

HELP ME PLEASE ILL MAKE U BRANLEIST IF WRITE

katen-ka-za [31]

so the 1st on is the one on the left, middle is right and the 3rd one is the right one

3 0

2 years ago

Read 2 more answers

The skateboarder has a mass of 100 kg. When traveling downward from E to D, he reaches a velocity of 11 m/s. Calculate his kinet

patriot [66]

6050 J is the kinetic energy at D

<u>Explanation:</u>

In physics, the object's kinetic energy (K.E) defined as the energy it possesses during movement. It can be defined as the required work to accelerate a certain body weight in order to rest at a certain speed. When the body receives this energy as it speeds up (accelerates), it retains this energy unless speed varies. The equation is given as,

$K . E=\frac{1}{2} \times m \times v^{2}$

Where,

m - mass of an object

v - velocity of the object

Here,

Given data:

m = 100 kg

v = 11 m/s

By substituting the given values in the above equation, we get

$K . E=\frac{1}{2} \times 100 \times(11)^{2}=\frac{1}{2} \times 100 \times 121=\frac{12100}{2}=6050\ \mathrm{J}$

6 0

3 years ago

There is a seasaw that's holding two men. The seesaw has a length of 18m that can pivot from a point at its center. Man 1 has a

Alex

Answer:

Distance= 2.3864m

Explanation:

So that the balance is in equilibrium parallel to the floor, we must match the moment each man makes with respect to the pivot point.

In many cases the point of application of force does not coincide with the point of application in the body. In this case the force acts on the object and its structure at a certain distance, by means of an element that transfers that action of this force to the object.

This combination of force applied by the distance to the point of the structure where it is applied is called the moment of force F with respect to the point. The moment will attempt a rotation shift or rotation of the object. The distance from the force to the point of application is called the arm.

Mathematically it is calculated by expression:

M= F×d

The moment caused by the first man is:

M1= 75kg × (9.81m/s²) × 1.75m= 1287.5625 N×m

The moment caused by the second man must be equal to that caused by the first by which:

M2= 1287.5625 N×m= 55kg × (9.81m/s²) × distance ⇒

⇒distance= (1287.5625 N×m)/( (55kg × (9.81m/s²) )= 2.3864m

At this distance from the pivot point, the second should sit down so that the balance is balanced parallel to the ground.

3 0

2 years ago

The force required to stretch a Hooke’s-law

Setler [38]

I think this is correct, but I am not entirely certain.

Find the force constant of the spring:

F = - KX

(0 - 62.4) = -K(0.172m)

-362.791 = -K

362.791 N/m = K

Find the work done in stretching the spring:

W = (1/2)KX

W = (1/2)(362.791)(0.172m)

W = 31.2 J

5 0

2 years ago