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sveta [45]
3 years ago
5

04-9 (ML 6.12) A pristine stream flowing at 100 cubic feet per second (cfs) in the Rocky Mountains contains 5 mg/L of suspended

solids (SS). During the spring, ice melt conveys an additional 250 mg/L of TSS at a rate of 20 cfs into the stream. Determine the concentration of TSS in the stream during the spring. (Hint: Treat the ice melt as a secondary stream added to the first.) Answer: 46 mg/L
Physics
1 answer:
timurjin [86]3 years ago
4 0

Answer:

C_{mix} = 45.833 mg/l

Explanation:

Give data:

flow rate of stream is Q_1 100 cfs

concentration of suspended solids SS_1 5 mg/L

During spring time,

concentration of TSS = SS_2 = 250 mg/l

flow rate of stream Q_2= 20 cfs

we know that, concentration of mix is given as

C_{mix} = \frac{Q_1 SS_1 +  Q_2 SS_2}{Q_1 +Q_2}

C_{mix} = \frac{100\times 5 + 20\times 250}{100+20}

C_{mix} = 45.833 mg/l

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Answer:

<h2>The answer is 12 m</h2>

Explanation:

The distance covered by an object given it's velocity and time taken can be found by using the formula

distance = velocity × time

From the question we have

distance = 2 × 6

We have the final answer as

<h3>12 m</h3>

Hope this helps you

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How can two machines appear identical and yet not have the same actual mechanical advantage
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If the magnetic flux through a loop increases according to the relation; where , is in milliweber (mWb) and t is in seconds. Cal
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The magnitude of e.m.f induced in the loop when t = 2 s is 31 Volts.

<h3>emf induced in the loop</h3>

The magnitude of e.m.f induced in the loop is calculated as follows;

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dФ/dt = 12t + 7

at t = 2 seconds

emf = dФ/dt = 12(2) + 7 = 31 V

Thus,  the magnitude of e.m.f induced in the loop when t = 2 s is 31 Volts.

Learn more about emf here: brainly.com/question/24158806

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6 0
2 years ago
A 91.0-kg hockey player is skating on ice at 5.50 m/s. another hockey player of equal mass, moving at 8.1 m/s in the
never [62]

The momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

(m₁) is the mass of hockey player 1= 91.0-kg

(m₂) is the mass of hockey player 2=  91.0-kg

(u₁) is the velocity before collision of hockey player 1 = 5.50 m/s.

(u₂) is the velocity before the collision of hockey player 2=?

a)

Momentum before the collision;

\rm  m_1u_1 + m_2u_2 \\\\ 91.0 \times 5.50 + 91.0 \times 8.1 \\\\ 1237.6 kg m/s^2

Momentum before the collision = 1237.6 kg m/s².

b)

The velocity of the two hockey players after the collision from the law of conservation of the momentum as:

Momentum before collision = Momentum after the collision

1237.6 kg m/s² = (m₁+m₂)V

1237.6 kg m/s² =(2 ×91.0-kg )V

V=6.8 m/sec.

Hence, momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

To learn more about the law of conservation of momentum refer;

brainly.com/question/1113396

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8 0
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