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JulijaS [17]
3 years ago
13

A village wishes to measure the quantity of water that is piped to a factory during a typical morning. A gauge on the water line

gives the flow rate (in cubic meters per hour) at any instant. The flow rate is about 100 m3/hr at 6 am and increases steadily to about 280 m3/hr at 9 am. Using only this information, give your best estimate of the total volume of water used by the factory between 6 am and 9 am.
Mathematics
1 answer:
anzhelika [568]3 years ago
3 0

Answer:

570 m³

Step-by-step explanation:

The volume of water is the product of flow rate of water and the time taken. We are to get the volume of water used between 6 am and 9 am, that is for 3 hours (9 - 6).

We are given the flow rate at 6 am and the flow rate at 9 am, but this flow rate changes between 6 am and 9 am. To get the estimate of the water used, Let us assume that it flows at the same flow rate as it was at 6 am throughout, hence:

V_L= 100 m^3/h * 3h=300\ m^3

Also, let us assume that it flows at the same flow rate as it was at 9 am throughout, hence:

V_R= 280 m^3/h * 3h=840\ m^3

To get the best estimate of the total volume, let us find the average of the two values, hence:

Volume=\frac{V_L+V_R}{2} =\frac{300+840}{2}=570\ m^3

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The height of this triangle would be 10.4

In order to find this, you first must find the length of the sides. Using a manipulated formula for area of an equilateral triangle, we can determine the lengths of the side. Below if the formula.

S = \frac{2}{3}3^{\frac{3}{4}} \sqrt{A}

In this, S is the length of the side and A is the area. So we plug in and get:

S = \frac{2}{3}3^{\frac{3}{4}} \sqrt{62.4}

S = \frac{2}{3}3^{\frac{3}{4}} 7.89

S = 12

Now that we have the side as 12, we can use the Pythagorean Theorem to find the height. If you split a equilateral triangle down the middle, you are left with two right triangles. Using this right triangle, the hypotenuse would be 12, the first leg would be 6 (half of the base) and the height would be the other leg. So we plug in and solve.

a^{2} + b^{2} = c^{2}

6^{2} + h^{2} = 12^{2}

36 + h^{2} = 144

h^{2} = 108

h = 10.4

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3 years ago
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Step-by-step explanation:

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NASA launches a rocket at t = 0 seconds. Its height, in meters above sea-level, as a function of time is given by h ( t ) = − 4.
Aleks [24]

Answer:

A.) 24.08 seconds

B.) 825.42 metres

Step-by-step explanation:

function of time is given as

h ( t ) = − 4.9 t 2 + 118 t + 115 .

Where a = -4.9, b = 118, c = 115

Let's assume that the trajectory of the rocket is a perfect parabola.

The time t the rocket will reach its maximum height will be at the symmetry of the parabola.

t = -b/2a

Substitute b and a into the formula

t = -118/-2(4.9)

t = 118/9.8

t = 12.041 seconds

Since NASA launches the rocket at t = 0 seconds, the time it will splash down into the ocean will be 2t.

2t = 2 × 12.041 = 24.08 seconds

Therefore, the rocket splashes down after 24.08 seconds.

B.) At maximum height, time t = 12.041s

Substitute t for 12.041 in the function

h ( t ) = − 4.9 t 2 + 118 t + 115

h(t) = -4.9(12.041)^2 + 118(12.041) + 115

h(t) = -4.9(144.98) + 118(12.041) + 115

h(t) = -710.402 + 1420.82 + 115

h(t) = 825.42 metres

Therefore, the rocket get to the peak at 825.42 metres

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3 years ago
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