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Mrac [35]
2 years ago
5

A swimming pool contains x (less than 0.02) grams of chlorine per cubic meter. the pool measures 5 meters by 50 meters and is 2

meters deep. some water will be drained and replaced by water contain 5 grams of chlorine per cubic meter. how much water should be drained so the pool ends up with 0.02 grams of chlorine per cubic meter.
Physics
1 answer:
zubka84 [21]2 years ago
3 0
The solution for this problem would be:(10 - 500x) / (5 - x) 
so start by doing: 
x(5*50*2) - xV + 5V = 0.02(5*50*2) 
500x - xV + 5V = 10 
V(5 - x) = 10 - 500x 
V = (10 - 500x) / (5 - x) 
(V stands for the volume, but leaves us with the expression for x)
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A 1900 kg car rounds a curve of 55 m banked at an angle of 11 degrees? . If the car is traveling at 98 km/h, How much friction f
Nat2105 [25]

Answer:

22000 N

Explanation:

Convert velocity to SI units:

98 km/h × (1000 m / km) × (1 h / 3600 s) = 27.2 m/s

Draw a free body diagram.  There are three forces acting on the car.  Normal force perpendicular to the bank, gravity downwards, and friction parallel to the bank.

I'm going to assume the friction force is pointed down the bank.  If I get a negative answer, that'll just mean it's actually pointed up the bank.

Sum of the forces in the radial direction (+x):

∑F = ma

N sin θ + F cos θ = m v² / r

Sum of the forces in the y direction:

∑F = ma

N cos θ - F sin θ - W = 0

To solve the system of equations for F, first solve for N and substitute.

N = (W + F sin θ) / cos θ

Substituting:

((W + F sin θ) / cos θ) sin θ + F cos θ = m v² / r

(W + F sin θ) tan θ + F cos θ = m v² / r

W tan θ + F sin θ tan θ + F cos θ = m v² / r

W tan θ + F (sin θ tan θ + cos θ) = m v² / r

W tan θ + F sec θ = m v² / r

F sec θ = m v² / r - W tan θ

F = m v² cos θ / r - W sin θ

F = m (v² cos θ / r - g sin θ)

Given that m = 1900 kg, θ = 11°, v = 27.2 m/s, and r = 55 m:

F = 1900 ((27.2)² cos 11 / 55 - 9.8 sin 11)

F = 21577 N

Rounding to two sig-figs, you need at least 22000 N of friction force.

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If it is completely elastic, you can calculate the velocity of the second ball from the kinetic energy 
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<span>v2' = velocity of #2 after collision. </span>

<span>kinetic energy: v1^2 = v1' ^2 + v2' ^2 (1/2 and m cancel out) </span>
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Suppose that during any period of ¼ second there is one instant at which the crests or troughs of component waves are exactly in
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<h3><u>Answer;</u></h3>

A. 4

<h3><u>Explanation;</u></h3>
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