During the storm a tree fell over into a river what might happen to the tree

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

How do you find the water pressure at the bottom of the 55-m-high water tower?

svp [43]

-- What's the volume of a cylinder with radius=1m and height=55m ?

   ( Volume of a cylinder = π R² h )

-- How much does that volume of water weigh ?

    1 liter of water = 1 kilogram of mass
Weight = (mass) x (acceleration of gravity)

-- What's the area of the bottom of that 1m-radius cylinder ?

   Pressure = (force) / (area)

5 0

3 years ago

An object is attached to a hanging unstretched, ideal and massless spring and slowly lowered to its equilibrium position, a dist

Ad libitum [116K]

Answer:

10.6cm

Explanation:

We are given 5.3cm below the starting point (spring extension).

Therefore, to find static vertical equilibrium, we use the equation:

kx = mg

Where:

k = spring constant =

=mg/5.3 kg/s²

We are told the object was dropped from rest.

Therefore:

loss in potential energy = gain in spring p.e

Let's use the expression:

mgx = ½kx²

We are asked to find the stretch at maximum elongation x.

To find x, we make x subject of the formula.

Therefore, we have:

x = 2mg/k (after rearranging the equation above)

x = (2mg) / (mg/5.3)

x = 10.6cm

3 0

2 years ago

Dolphins communicate underwater using sound waves. This is called ________.

Ray Of Light [21]

Answer:

D. echolocation

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7 0

2 years ago

Read 2 more answers

In a car lift used in a service station, compressed air exerts a force on a small piston that has a circular cross section and a

timofeeve [1]

Answer:

a) F₁ = 1.48 x 10³ N

b) P = 1.88*10⁵ Pa

c) The work is equal in both pistons

Explanation:

Pascal´s Principle can be applied in the hydraulic press:

If we apply a small force (F₁) on a small area piston (A₁), then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F₂) can be exerted that is proportional to the area(A₂) of the piston.

Pressure is defined as the force per unit area:

$P=\frac{F}{A}$ Formula (1)

P₁=P₂

$\frac{F_{1} }{A_{1} } =\frac{F_{2} }{A_{2} }$ Formula (2)

Data

r₁= 5 cm = 0.05 m

r₂= 15 cm = 0.15 m

F₂= 13300N

Area of the pistons (A₁,A₂)

A=π*r² : Area of the circle

A₁ = π*(0.05)²=7.85*10⁻³ m²

A₂= π*(0.15)²= 70.69*10⁻³ m²

a) Force that compressed air must exert to lift a car weighing 13300 N

We replace data in the formula (2)

$\frac{F_{1} }{A_{1} } =\frac{F_{2} }{A_{2} }$

$F_{1} = \frac{13300*7.85*10^{-3} }{70.69*10^{-3} }$

F₁ = 1.48 x 10³ N

b) Air pressure produced by F₁

We replace data in the formula (1)

$P=\frac{F}{A}$

F₁ = 1.48 x 10³ N , A₁ = 7.85*10⁻³ m²

$P=\frac{1.48*10^{3} }{7.85*10^{-3} }$

P= 1.88*10⁵ Pa

c)The volume of liquid displaced by the small piston is distributed in a thin layer on the large piston, so that the product of the force by the displacement (the work) is equal in both pistons.

3 0

3 years ago