Ask question

Ask question

All categories

3 years ago

11

To aid in the prevention of tooth decay, it is recommended that drinking water contain 0.800 ppm fluoride, F−. How many grams of

F− must be added to a cylindrical water reservoir having a diameter of 2.02 × 102 m and a depth of 87.32 m?

1 answer:

Leviafan [203]3 years ago

4 0

Answer:

2.23 × 10^6 g of F- must be added to the cylindrical reservoir in order to obtain a drinking water with a concentration of 0.8ppm of F-

Explanation:

Here are the steps of how to arrive at the answer:

The volume of a cylinder = ((pi)D²/4) × H

Where D = diameter of the cylindrical reservoir = 2.02 × 10^2m

H = Height of the reservoir = 87.32m

Therefore volume of cylindrical reservoir = (3.142×202²/4)m² × 87.32m = 2798740.647m³

1ppm = 1g/m³

0.8ppm = 0.8 × 1g/m³

= 0.8g/m³

Therefore to obtain drinking water of concentration 0.8g/m³ in a reservoir of volume 2798740.647m³, F- of mass = 0.8g/m³ × 2798740.647m³ = 2.23 × 10^6 g must be added to the tank.

Thank you for reading.

You might be interested in

Is it possible to produce a continuous and oriented aramid fiber–epoxy matrix composite having longitudinal and transverse modul

gregori [183]

Answer:

It is not possible to produce continued and oriented fiber.

Explanation:

To solve the problem it is necessary to take into account the concepts related to Fiber volume ratio. The amount of fiber in a fiber reinforced compound corresponds directly to the mechanical properties of the compound. Given the fiber volume fraction, the theoretical elastic properties of a compound can be determined. The elastic modulus of a compound in the fiber direction of a unidirectional compound can be calculated using the following equation:

$E = (1-V_f)E_m+V_fE_f$

Where,

E is the longitudinal modulus of Elasticity

$V_f$ is the fiber volume ratio

$E_m$ is the elastic modulus of the matrix

$E_f$ is the elastic modulus of the fibers

We need to consult the table of characteristics of Fibers and Reinforcements of Materials, in which they specify that the modulus of elasticity of the aramid fiber-epoxy is

$E_f = 131Gpa$

Moreover from the statement,

$E = 35Gpa$

$E_m = 3.4Gpa$

Replacing in the previous equation,

$35 = 3.4 (1-V_f)+131V_f$

$V_f = 0.25 \rightarrow longitudinal$

To make the comparison we now calculate the Fiber volume ratio through the transverse elastic modulus,

$E = \frac{E_mE_f}{(1-V_f)E_f+V_fE_f}$

Our values are given in this case as:

$E = 5.17Gpa\\E_m = 3.4Gpa \\E_f = 131Gpa$

Replacing,

$5.17 = \frac{3.4*131}{(1-V_f)(131)+V_f*3.4}$

$V_f = 0.351 \rightarrow transversal$

From both cases it is possible to conclude that it is not possible to produce a fiber of the specified material in a continuous and oriented manner, as long as the volume fraction is different in the different cases.

5 0

3 years ago

A 2 kg block slides on a rough horizontal surface with muk=0.6. It has an initial velocity of 5 m/s. Use g = 10 m/s2

Irina18 [472]

Answer:

360000

Explanation:

4 0

3 years ago

I'm feeling nice so.... F.r.e.e POINTS (pls follow me)

NISA [10]

Answer:

ty :,)

Explanation:

6 0

3 years ago

Read 2 more answers

A dielectric-filled parallel-plate capacitor has plate area A = 30.0 cm2 , plate separation d = 9.00 mm and dielectric constant

PIT_PIT [208]

Answer:

$9.96\cdot 10^{-10}J$

Explanation:

The capacitance of the parallel-plate capacitor is given by

$C=\epsilon_0 k \frac{A}{d}$

where

ϵ0 = 8.85x10-12 C2/N.m2 is the vacuum permittivity

k = 3.00 is the dielectric constant

$A=30.0 cm^2 = 30.0\cdot 10^{-4}m^2$ is the area of the plates

d = 9.00 mm = 0.009 m is the separation between the plates

Substituting,

$C=(8.85\cdot 10^{-12}F/m)(3.00 ) \frac{30.0\cdot 10^{-4} m^2}{0.009 m}=8.85\cdot 10^{-12} F$

Now we can calculate the energy of the capacitor, given by:

$U=\frac{1}{2}CV^2$

where

C is the capacitance

V = 15.0 V is the potential difference

Substituting,

$U=\frac{1}{2}(8.85\cdot 10^{-12}F)(15.0 V)^2=9.96\cdot 10^{-10}J$

4 0

3 years ago

A student is examining a bacterium under the microscope. The E. coli bacterial cell has a mass of m = 0.900 fg (where a femtogra

just olya [345]

Answer:

9.73 x 10⁻¹⁰ m

Explanation:

According to Heisenberg uncertainty principle

Uncertainty in position x uncertainty in momentum ≥ h / 4π

Δ X x Δp ≥ h / 4π

Δp = mΔV

ΔV = Uncertainty in velocity

= 2 x 10⁻⁶ x 3 / 100

= 6 x 10⁻⁸

mass m = 0.9 x 10⁻¹⁵ x 10⁻³ kg

m = 9 x 10⁻¹⁹

Δp = mΔV

= 9 x 10⁻¹⁹ x 6 x 10⁻⁸

= 54 x 10⁻²⁷

Δ X x Δp ≥ h / 4π

Δ X x 54 x 10⁻²⁷ ≥ h / 4π

Δ X = h / 4π x 1 / 54 x 10⁻²⁷

= $\frac{6.6\times10^{-34}}{4\times3.14\times54\times10^{-27}}$

= 9.73 x 10⁻¹⁰ m

7 0

3 years ago