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arsen [322]
3 years ago
7

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

ce as expensive as laying it on land. What path should the pipe follow in order to minimize the cost?
Physics
1 answer:
shusha [124]3 years ago
8 0

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

\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.

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Decreasing the distance between two objects having a considerable mass would increase the attraction on gravitation. The reverse is true that if you separate or inrease the objects distance would substantially decrease their gravitational attraction. Most object in our planet is held by its gravitational force towards it's center.
5 0
3 years ago
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A trampoline spring has a force constant k = 800 N/m and is stretched exactly 17.5cm. What is the energy required to do this?
Artist 52 [7]

Answer:

the energy required for the extension is 12.25 J

Explanation:

Given;

force constant of trampoline spring, k = 800 N/m

extension of trampoline spring, x = 17.5 cm = 0.175 m

The energy required for the extension is calculated as;

E = ¹/₂kx²

E = 0.5 x 800 x 0.175²

E = 12.25 J

Therefore, the energy required for the extension is 12.25 J

6 0
3 years ago
What is Newton's Second Law of Motion? (2 points) Every reaction is equal to the force applied. Forces are balanced when they ar
raketka [301]

Well, they're not quite the way Newton expressed it, but out of all this mess of statements, there are two that are correct AND come from Newton's 2nd Law of Motion:

<em>-- The smaller the mass of an object, the greater the acceleration of that object when a force is applied. </em>

<em>-- The greater the force applied, the greater the acceleration.</em>

For the <u><em>other </em></u>statements in the question:

-- <em>Every reaction is equal to the force applied.</em>  True; comes from Newton's <u><em>3rd</em></u> law of motion.

-- <em>Forces are balanced when they are equal and opposite.</em>  True; kind of a definition, not from Newton's laws of motion.

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5 0
3 years ago
A small 1.0 kg steel ball rolls west at 3.0 m/s collides with a large 3.0 kg ball at rest. After the collision, the small ball m
77julia77 [94]

Answer:

The direction of the momentum of the large ball after the collision with respect to east is 146.58°.

Explanation:

Given that,

Mass of large ball = 3.0 kg

Mass of steel ball = 1.0 kg

Velocity = 3.0 kg

After collision,

Velocity = 2.0 m/s

Using conservation of momentum

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

3.0\times0+1.0\times(3.0)(-i)=1.0\times2(-j)+3.0\times v_{2}

-3i+2j=3.0\times v_{2}

v_{2}=-i+0.66j

The direction of the momentum

tan\theta=\dfrac{0.66}{-1}

\theta=tan^{-1}\dfrac{0.66}{-1}

\theta=-33.42^{\circ}

The direction of the momentum with respect to east

\theta=180-33.42=146.58^{\circ}

Hence, The direction of the momentum of the large ball after the collision with respect to east is 146.58°.

7 0
3 years ago
Find the total electric charge of 2.5 kg of electrons. Express your answer using two significant figures.
zimovet [89]

Answer : The total electric charge of electrons is, -4.4\times 10^{11}C

Explanation:

Answer : The number of electrons transferred are, 4.68\times 10^{20}

Explanation :

First we have to calculate the number of electrons.

Number of electrons = \frac{\text{Total mass of electrons}}{\text{Mass of one electron}}

Mass of 1 electron = 9.1\times 10^{-31}kg

Total mass of electron = 2.5 kg

Number of electrons = \frac{2.5kg}{9.1\times 10^{-31}kg}

Number of electrons = 2.75\times 10^{30}

Now we have to calculate the total electric charge of electrons.

Formula used :

Q=ne\\\\n=\frac{Q}{e}

where,

n = number of electrons transferred = 2.75\times 10^{30}

Q = charge on electrons = ?

e = charge on 1 electron = -1.602\times 10^{-19}C

Now put all the given values in the above formula, we get:

2.75\times 10^{30}=\frac{Q}{-1.602\times 10^{-19}C}

Q=-4.4\times 10^{11}C

Thus, the total electric charge of electrons is, -4.4\times 10^{11}C

4 0
3 years ago
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