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Ainat [17]
2 years ago
13

How are humans potentially vulnerable to some of these same chemicals? How are humans potentially vulnerable to some of these sa

me chemicals?
What should the government and/or industry do about these emerging contaminants?

What can you do to make water safer for yourself, your community, and the fish?
Physics
1 answer:
Svetradugi [14.3K]2 years ago
6 0
The 5 major unit processes include chemical coagulation, flocculation, sedimentation, filtration, and disinfection (described below). There are chemicals added to the water as it enters the various treatment processes.
You might be interested in
A body of mass 25kg, moving at 3 ms per second on a rough horizontal floor brought to rest after sliding through a distance of 2
erastova [34]
You have to solve this by using the equations of motion:
u=3
v=0
s=2.5
a=?
v^2=u^2+2as
0=9+5s
Giving a=-1.8m/s^2

Then using the equation:
F=ma
F is the frictional force as there is no other force acting and its negative as its in the opposite direction to the direction of motion.

-F=25(-1.8)
F=45N

Then use the formula:
F=uR
Where u is the coefficient of friction, R is the normal force and F is the frictional force.

45=u(25g)
45=u(25*10)

Therefore, the coefficient of friction is 0.18

Hope that helps




5 0
3 years ago
Which of these is an example of acceleration ?
anygoal [31]

Answer:

no picture or anything so cant anwser

Explanation:

8 0
3 years ago
Read 2 more answers
At the moment a hot cake is put in a cooler, the difference between the cakes and the cooler's
hram777 [196]

Answer: D(t)= 50(4/5)^t

Explanation: If 1/5 of the temperature difference is lost each minute, that means 4/5 of the difference remains each minute. So each minute, the temperature difference is multiplied by a factor of 4/5 (or 0.8).

If we start with the initial temperature difference, 50° Celsius, and keep multiplying by 4/5, this function gives us the temperature difference t minutes after the cake was put in the cooler.

5 0
3 years ago
Read 2 more answers
A man is standing on a weighing machine on a ship which is bobbing up and down with simple harmonic motion of period T=15.0s.Ass
STALIN [3.7K]

Well, first of all, one who is sufficiently educated to deal with solving
this exercise is also sufficiently well informed to know that a weighing
machine, or "scale", should not be calibrated in units of "kg" ... a unit
of mass, not force.  We know that the man's mass doesn't change,
and the spectre of a readout in kg that is oscillating is totally bogus.

If the mass of the man standing on the weighing machine is 60kg, then
on level, dry land on Earth, or on the deck of a ship in calm seas on Earth,
the weighing machine will display his weight as  588 newtons  or as 
132.3 pounds.  That's also the reading as the deck of the ship executes
simple harmonic motion, at the points where the vertical acceleration is zero.

If the deck of the ship is bobbing vertically in simple harmonic motion with
amplitude of M and period of 15 sec, then its vertical position is 

                                     y(t) = y₀ + M sin(2π t/15) .

The vertical speed of the deck is     y'(t) = M (2π/15) cos(2π t/15)

and its vertical acceleration is          y''(t) = - (2πM/15) (2π/15) sin(2π t/15)

                                                                = - (4 π² M / 15²)  sin(2π t/15)

                                                                = - 0.1755 M sin(2π t/15) .

There's the important number ... the  0.1755 M.
That's the peak acceleration.
From here, the problem is a piece-o-cake.

The net vertical force on the intrepid sailor ... the guy standing on the
bathroom scale out on the deck of the ship that's "bobbing" on the
high seas ... is (the force of gravity) + (the force causing him to 'bob'
harmonically with peak acceleration of  0.1755 x amplitude).

At the instant of peak acceleration, the weighing machine thinks that
the load upon it is a mass of  65kg, when in reality it's only  60kg.
The weight of 60kg = 588 newtons.
The weight of 65kg = 637 newtons.
The scale has to push on him with an extra (637 - 588) = 49 newtons
in order to accelerate him faster than gravity.

Now I'm going to wave my hands in the air a bit:

Apparent weight = (apparent mass) x (real acceleration of gravity)

(Apparent mass) = (65/60) = 1.08333 x real mass.

Apparent 'gravity' = 1.08333 x real acceleration of gravity.

The increase ... the 0.08333 ... is the 'extra' acceleration that's due to
the bobbing of the deck.

                        0.08333 G  =  0.1755 M

The 'M' is what we need to find.

Divide each side by  0.1755 :          M = (0.08333 / 0.1755) G

'G' = 9.0 m/s²
                                       M = (0.08333 / 0.1755) (9.8) =  4.65 meters .

That result fills me with an overwhelming sense of no-confidence.
But I'm in my office, supposedly working, so I must leave it to others
to analyze my work and point out its many flaws.
In any case, my conscience is clear ... I do feel that I've put in a good
5-points-worth of work on this problem, even if the answer is wrong .

8 0
3 years ago
A 20.0 kg crate sits at rest at the bottom of a 15.0-m-long ramp that is inclined at 34.0° above horizontal. A constant horizont
ankoles [38]

Answer:

987 joules, 3.01s

Explanation:

(A)

from the attached diagram

net force, Fnet, pulling the crate up the ramp is given by

Fnet = FcosФ - WsinФ - Fr

where FcosФ is the component of horizontal force 290N resolved parallel to the plane

WsinФ = mgsinФ = component of the weight of the crate resolved parallel to the plane

Fr = constant opposing frictional force

Fnet = 290cos34⁰ - 20 × 9.8 × sin34° - 65

Fnet = 240.421 - 109.602 - 65

Fnet = 65.82N

Work done on the crate up the ramp, W, is given by

W = Fnet × d (distance up the plane)

W = 65.819 × 15

W = 987.285 joules

W = 987 joules (to 3 significant Figures)

(B)

to calculate the time of travel up the ramp

we use the equation of motion

s = ut + \frac{1}{2}at^{2}

where s = distance up the plane, 15m

u = Initial velocity of the crate, which is 0 for a body that is initially at rest

a = acceleration up the plane, given by

a = \frac{Fnet}{m}

where m = mass of the crate, 20 kg

now, a = \frac{65.819}{20} \\a = 3.291\frac{m^{2} }{s}

from, s = ut + \frac{1}{2}at^{2}

15 = 0*t + \frac{1}{2}* 3.291 * t^{2}

15 = 0 + 1.645t^{2}

15 = 1.645t^{2}

t = \sqrt{\frac{15}{1.645} }

t = 3.019

t = 3.01s (to 3 sig fig)

7 0
3 years ago
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