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3 years ago

A groundwater contains the following cations (expressed as the cation):

Engineering

2 answers:

Hunter-Best [27]3 years ago

6 0

Answer:

208 mg/L

Explanation:

Only Mg++ and Ca++ causes hardness not Na+

Given

Mg++ = 20 mg/L

Ca++ = 50 mg/L

$Hardness\ of\ water = C^{2+}\times \frac{equivalent\ weight\ of\ CaCO3}{equivalent\ mass\ of\ C^{2+}}\ +\ mg^{2+}\times \frac{equivalent\ weight\ of\ CaCO3}{equivalent\ mass\ of\ mg^{2+}}$

= $50\times\frac{50}{20}\ +\ 20\times\frac{50}{12}$

= 208.33 mg/L $\approx$ 208mg/L

Oliga [24]3 years ago

5 0

Answer:

50mg/L

Explanation:

CaCO3 -> Ca++ + CO2

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Can a 1½ " conduit, with a total area of 2.04 square inches, be filled with wires that total 0.93 square inches if the maximum f

Papessa [141]

Answer:

it is not possible to place the wires in the condui

Explanation:

given data

total area = 2.04 square inches

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Can it is possible place wire in conduit conduit

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we know maximum fill is 40%

so here first we get total area of conduit that will be

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total area of conduit = 0.816 square inches

but this area is less than required area of wire that is 0.93 square inches

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3 years ago

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MrMuchimi

Answer:

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Explanation:

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3 years ago

The lead of the threaded shaft of the C-clamp is 0.05 in., and the mean radius of the thread is r = 0.15 in. Assume mus = 0.18.

barxatty [35]

The omitted part of the question shown in bold format

The lead of the threaded shaft of the C-clamp is 0.05 in., and the mean radius of the thread is r = 0.15 in. Assume μs = 0.18 and μk = 0.16. What couple must be applied to the shaft to exert a 30-lb force on the clamped object?

Answer:

C = 0.967 in. lb

Explanation:

Given that:

The lead of the threaded shaft of the C-clamp is 0.05 in.

∴ the pitch of the screw = 0.05 in

the mean radius of the thread is r = 0.15 in.

Assuming:

(μs)= 0.18 which implies the coefficient of the static friction

(μk) = 0.16 (coefficient of kinetic friction)

Force = 30-lb

What couple must be applied to the shaft to exert a 30-lb force on the clamped object?

To determine the Couple (C) that must be applied; we use the expression:

C = Fr tan ( $\theta_k$ + $\alpha$ )

where; F = force

r = mean radius

$\theta_k$ = angle of kinetic friction

$\alpha$ = pitch angle

NOW, let's take then one after the other.

From the coefficient of the static friction and the kinetic friction; we can solve for their respective angles, so we have:

Angle of static friction ( $\theta_s$ ) $= tan^{-1}(U_s)$

( $\theta_s$ ) = $tan^{-1}(0.18)$

( $\theta_s$ ) = 10.204°

Angle of kinetic friction ( $\theta_k$ ) $= tan^{-1}(U_k)$

$\theta_k$ $= tan^{-1}(0.16)$

$\theta_k$ = 9.0903°

To determine the pitch angle( $\alpha$ ); we apply the expression:

( $\alpha$ ) = $tan^{-1}(\frac{p}{2 \pi r} )$

( $\alpha$ ) = $tan^{-1}(\frac{0.05}{2 \pi 0.15} )$

( $\alpha$ ) = $tan^{-1}(0.0530516 )$

( $\alpha$ ) = 3.0368°

Have gotten our parameters to solve for Couple (C); then we have:

C = Fr tan ( $\theta_k$ + $\alpha$ )

substituting our values; we have:

C = (30 × 0.15) tan ( 9.0903 + 3.0368)

C = 4.5 × tan ( 12.1271)

C = 4.5 × 0.2148761968

C = 0.96669428854 in.lb

C = 0.967 in. lb

Therefore, 0.967 in. lb couple must be applied to the shaft to exert a 30-lb force on the clamped object.

6 0

4 years ago

Why water parameters of Buriganga river vary between wet and dry seasons?<br> Explain.

Sedaia [141]

I honestly don’t know because I honestly don’t know what I mean

5 0

3 years ago

Consider the same piping system. this time, the same pipe is buried underground. assuming that there is a constant heat flux of

charle [14.2K]

Complete Question

Complete Question is attached below

Answer:

Option A

Explanation:

From the question we are told that:

inner Diameter of pipe $d_i=100^c$

Thickness $t=50mm$

Outer diameter of pipe $d_o=1.1m$

Length $l=5m$

Temperature $T_i=130^oC$

Generally the equation for Heat Balance is mathematically given by

$q*\pi d_oL=mC_p(T_i-T_o)$

Therefore

$T_o=T_i+\frac{q*\pi d_oL}{mC_p}$

$T_o=130+\frac{100*3.142 *1.1*5}{0.5*4000}$

$T_o=129.136^oC$

Therefore the exit temperature of the water.is $T_o=129.136^oC$

Option A

7 0

3 years ago