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Makovka662 [10]
3 years ago
5

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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ruslelena [56]
Usually the first digit of the vin id’s the country it was built. So technician A would be correct. That’s usually how it is. Hope this helps. Please let me know if this is incorrect
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3 years ago
1 A power transmission includes a belt drive, a chain drive and a gear drive. Which of the following is the best arrangement bet
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Answer:d

Explanation:

(d) chain drive belt drive gear drive

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3 years ago
Assume that a specific hard disk drive has an average access time of 16ms (i.e. the seek and rotational delay sums to 16ms) and
djyliett [7]

Answer:

Average access time for a hard disk = 11.38 ms

Explanation:

See attached pictures for step by step explanation.

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4 years ago
A transformer has 300,000 windings in its primary coil and uses 12,000V AC input. (4 points) How many windings would be needed t
viva [34]

Answer:

  2750

Explanation:

The number of windings and the voltage are proportional.

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Let n represent the number of windings to produce 110 Vac. Then the proportion is ...

  n/110 = 300,000/12,000

  n = 110(300/12) = 2750 . . . . multiply by 110

2750 windings would be needed to produce 110 Vac at the output.

7 0
2 years ago
A horizontal curve on a two-lane highway (10-ft lanes) is designed for 50 mi/h with a 6% superelevation. The central angle of th
fenix001 [56]

Answer:

The PT station is at 485+20.02 and 21.92 ft are to be cleared from the lane's shoulder to provide adequate stopping sight distance.

Explanation:

From table 3.5 of Traffic Engineering by Mannering

R_v=835

R=835+(10ft/2)= 840 ft.

Now T is given as

T=R tan(Δ/2)

Here Δ is the central angle of curve given as 35°

So

T=R tan(Δ/2)

T=840 x tan(35/2)

T=840 x tan(17.5)

T=264.85

Now

STA PC=482+72-(2+64.85)=480+07.15

Also L is given as

L=(π/180)RΔ

Here R is the radius calculated as 840 ft, Δ is the angle given as 35°.

L=(π/180)RΔ

L=(π/180)x840 x35

L=512.87 ft

STA PT=480+07.15+5+12.87=485+20.02

Now Ms is the minimum distance which is given as

M_s=R_v(1-cos(\frac{90 \times SSD}{\pi Rv}))\\

Here R_v is given as 835

SSD for 50 mi/hr is given as 425 ft from table 3.1 of Traffic Engineering by Mannering

So Ms is

M_s=R_v(1-cos(\frac{90 \times SSD}{\pi Rv}))\\M_s=835(1-cos(\frac{90 \times 425}{\pi 835}))\\M_s=26.92 ft

Now for the clearance from the inside lane

Ms=Ms-lane length

Ms=26.92-5= 21.92 ft.

So the PT station is at 485+20.02 and 21.92 ft are to be cleared from the lane's shoulder to provide adequate stopping sight distance.

6 0
3 years ago
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