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Mrac [35]
3 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]3 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)
You might be interested in
Charges q3 and q5 are now replaced by two charges, q2 and q6, having equal magnitude and sign (-3.4 μC). Charge q2 is located at
ivanzaharov [21]
I believe the answer would be zero because the q1 and q2 are equal on opposite sides and it

hope this helps
3 0
3 years ago
The drawing shows a large cube (mass = 21.0 kg) being accelerated across a horizontal frictionless surface by a horizontal force
MaRussiya [10]

Answer:

The blocks must be pushed with a force higher than 359 Newtons horizontally in order to accomplish this friction levitation feat.

Explanation:

The first step in resolving any physics problem is to draw the given scenario (if possible), see the attached image to have an idea of the objects and forces involved.

The large cube in red is being pushed from the left by a force \vec{P} whose value is to be found. That cube has its own weight \vec{w}_1=m_1\vec{g}, and it is associated with the force of gravity which points downward. Newton's third law stipulates that the response from the floor is an upward pointing force on the cube, and it's called the normal force \vec{N}_1.

A second cube is being pushed by the first, and since the force \vec{P} is strong enough it is able to keep such block suspended as if it were glued to the first one, due to friction. As in the larger cube, the smaller one has a weight \vec{w}_2=m_2\vec{g} pointing downwards, but the normal force in this block doesn't point upwards since its 'floor' isn't below it, but in its side, therefore the normal force directs it to the right as it is shown in the picture. Normal forces are perpendicular to the surface they contact. The final force is the friction between both cubes, that sets a resistance of one moving parallel the other. In this case, the weight of the block its the force pointing parallel to the contact surface, so the friction opposes that force, and thus points upwards. Friction forces can be set as Fr=\mu~N, where \mu is the coefficient of static friction between the cubes.

Now that all forces involved are identified, the following step is to apply Newton's second law and add all the forces for each block that point in the same line, and set it as equal its mass multiplied by its acceleration. The condition over the smaller box is the relevant one so its the first one to be analyzed.

In the vertical component: \Sigma F^2_y=Fr-w_2=m_2 a_y Since the idea is that it doesn't slips downwards, the vertical acceleration should be set to zero a_y=0, and making explicit the other forces: \mu N_2-m_2g=0\quad\Rightarrow (0.710)N_2-(4.5)(10)=0\quad\Rightarrow N_2=(4.5)(10)/(0.710)\approx 63.38 [N]. In the last equation gravity's acceleration was rounded to 10 [m/s^2].

In its horizontal component: \Sigma F^2_x=N_2=m_2 a_x, this time the horizontal acceleration is not zero, because it is constantly being pushed. However, the value of the normal force and the mass of the block are known, so its horizontal acceleration can be determined: 63.38=(4.5) a_x \quad \Rightarrow a_x=(63.38)/(4.5)\approx 14.08 [m/s^2]. Notice that this acceleration is higher than the one of gravity, and it is understandable since you should be able to push it harder than gravity in order for it to not slip.

Now the attention is switched to the larger cube. The vertical forces are not relevant here, since the normal force balances its weight so that there isn't vertical acceleration. The unknown force comes up in the horizontal forces analysis: \Sigma F_x=P=m a_x, since the force \vec{P} is not only pushing the first block but both, the mass involved in this equation is the combined masses of the blocks, the acceleration is the same for both blocks since they move together; P=(21.0+4.5) 14.08\approx 359.04 [N]. The resulting force is quite high but not impossible to make by a human being, this indicates that this feat of friction suspension is difficult but feasable.

4 0
3 years ago
True or False: There is the same amount of water on earth today as there was when earth was formed.
Nataliya [291]

Answer:

Yes the water amounts are the same.

Explanation:

4 0
3 years ago
Review. For a certain type of steel, stress is always proportional to strain with Young's modulus 20 × 10¹⁰ N/m² . The steel has
timurjin [86]

1.58\times 10^{-4}\ \mathrm{s}$ is the time interval elapses before the back end of the rod receives the message that it should stop.

Given:

Length of the rod, L = 80 cm = 0.800 m

Young's modulus, Y = 20 \times 10^{10}\;N/m^{2}

steel density, \rho = 7.86 \times 10^{3}\;kg/m^{3}

The speed of the wave in the rod is,

$v = \sqrt{\frac{Y}{\rho}} = \sqrt{\frac{20\times 10^{10}\ \mathrm{N/m^2}}{7.86\times 10^3\ \mathrm{kg/m^3}}} = 5044\ \mathrm{m/s}$

Consequently, the length of the rod's end is traveled by the wave in at

$t=\frac{L}{v} = \frac{0.800\ \mathrm{m}}{5044\ \mathrm{m/s}} = 1.58\times 10^{-4}\ \mathrm{s}$

Hence, 1.58\times 10^{-4}\ \mathrm{s}$ is the time interval elapses before the back end of the rod receives the message that it should stop.

<h3>What are Newtons Laws?</h3>

The three fundamental laws of classical mechanics known as Newton's laws of motion describe how an object's motion and the forces acting on it interact. The following paraphrase of these statutes is available

Unless a force acts upon a body, it remains at rest or in continual straight-line motion.

When a force acts on a body, the force is equal to the time rate at which the body's momentum changes.

When two bodies exert force on one another, the direction and amount of the force are opposed.

Isaac Newton first identified the three laws of motion in his 1687 book Philosophize Naturalis Principia Mathematica (Mathematical Principles of Natural Philosophy).

They served as the cornerstone for classical mechanics as Newton used them to examine and explain the motion of numerous physical objects and systems. The conceptual foundations of classical physics have been reconstructed in several ways since Newton, utilizing various mathematical techniques that have revealed insights that were hidden in the original, Newtonian formulation.

To know more about Newtons Laws, visit:

brainly.com/question/27573481

#SPJ4

7 0
1 year ago
Oil (SAE 30) at 15.6 oC flows steadily between fixed, horizontal, parallel plates. The pressure drop per unit length along the c
solong [7]

Answer:

q = 0.0003649123 m²/s = (3.65 × 10⁻⁴) m²/s

Explanation:

For laminar flow between two parallel horizontal plates, the volumetric flow per metre of width is given as

q = (2h³/3μ) (ΔP/L)

h = hydraulic depth = 4mm/2 = 2mm = 0.002 m

μ = viscosity of oil (SAE 30) at 15.6°C = 0.38 Pa.s

(ΔP/L) = 26 KPa/m = 26000 Pa/m

q = (2h³/3μ) (ΔP/L)

q = (26000) × (2(0.002³)/(3×0.38))

q = 0.0003649123 m²/s = (3.65 × 10⁻⁴) m²/s

5 0
3 years ago
Read 2 more answers
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