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3 years ago

What type of pollution did the Clean Water Act succeed in limiting?

Physics

2 answers:

Ilya [14]3 years ago

4 0

Sewage. If thats not it, then I need to see your choices. :)

lawyer [7]3 years ago

4 0

sewage is the answer because it's what the water was filled with most

You might be interested in

A rectangular box has side 10cm x 5cm x 2cm. If it lies on a horizontal surface and has weight 1.00N, calculate the maximum and

Vladimir [108]

Answer:

P max = 1000 pa

P min = 200 pa

Explanation:

P = F/A

pressure will be maximum when surface gets minimum. so to find the maximum amount of pressure we need to calculate the minimum surface. it is 2cm×5cm = 10cm² = 0.001m² . then we have:

P = 1 / 0.001 = 1000 pa

to find minimum pressure the surface that must be chosen is 10cm×5cm = 50cm² = 0.005m² .

P = 1 / 0.005 = 200 pa

7 0

2 years ago

Please need help fast

iVinArrow [24]

(a) See graph in attachment

The appropriate graph to draw in this part is a graph of velocity vs time.

In this problem, we have a horse that accelerates from 0 m/s to 15 m/s in 10 s.

Assuming the acceleration of the horse is uniform, it means that the velocity (y-coordinate of the graph) must increase linearly with the time: therefore, the velocity-time graph will appear as a straight line, having the final point at

t = 10 s

v = 15 m/s

(b) $1.5 m/s^2$

The average acceleration of the horse can be calculated as:

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time interval

In this problem,

u = 0

v = 15 m/s

t = 10 s

Substituting,

$a=\frac{15-0}{10}=1.5 m/s^2$

(c) 75 m

For a uniformly accelerated motion, the distance travelled can be calculated by using the suvat equation:

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

where

s is the distance travelled

u is the initial velocity

t is the time interval

a is the acceleration

In this problem,

u = 0

t = 10 s

$a=1.5 m/s^2$

Substituting,

$s=0+\frac{1}{2}(1.5)(10)^2=75 m$

(d) See attached graphs

In a uniformly accelerated motion:

- The distance travelled (x) follows the equation mentioned in part c,

$x=ut+\frac{1}{2}at^2$

So, we see that this has the form of a parabola: therefore, the graph x vs t will represents a parabola.

- The acceleration is constant during the motion, and its value is

$a=1.5 m/s^2$ (calculated in part b)

therefore, the graph acceleration vs time will show a flat line at a constant value of 1.5 m/s^2.

6 0

3 years ago

Suppose that you have a rod that is positively charged, a rod that is negatively charged and a stand that allows rotation. If yo

amm1812

Answer:

Explanation:

A charge is produced when an atom losses or gains an electron. The law of static electricity states that like charges repels, while unlike charges attracts.

1. To determine the charge on the polystyrene rod.

Place the polystyrene rod on the non-conducting rotating stand, and bring the positively charged rod close to it. If attraction occurs, it shows that it is oppositely charged. If repulsion occurs, it shows that it is positively charged.

Bringing a negatively charged rod close to the rotating polystyrene rod would attract it if the charge is opposite. But if the charge on the two rods are the same, repulsion occurs.

2a. When the polystyrene rod is positively charged, it would attract the negatively charged rod but repel the positively charged rod.

b. When the polystyrene rod is negatively charged, it would repel the negatively charged rod but attract the positively charged rod.

c. When the polystyrene rod is uncharged, no reaction would be observed when either the positively charged or negatively charged rod is brought close to it.

3 0

3 years ago

When a car of mass 1167 kg accelerates from 10.0 m/s to some final speed, 4.00  10 5J of work are done. Find this final speed.

Akimi4 [234]

Mass=1167kg
Initial velocity=u=10m/s
Acceleration=a=4m/s^2
Work done=105J=W
Final velocity=v=?
Force=F
Distance=d

Apply Newton's second law

$\\ \tt\hookrightarrow F=ma$

$\\ \tt\hookrightarrow F=1167(4)=4668N$

Now

$\\ \tt\hookrightarrow W=Fd$

$\\ \tt\hookrightarrow d=\dfrac{W}{F}$

$\\ \tt\hookrightarrow d=\dfrac{105}{4668}$

$\\ \tt\hookrightarrow d=0.022m$

Now

d be s

According to third equation of kinematics

$\\ \tt\hookrightarrow v^2=u^2+2as$

$\\ \tt\hookrightarrow v^2=10^2+2(4)(0.022)$

$\\ \tt\hookrightarrow v^2=100+8(0.022)$

$\\ \tt\hookrightarrow v^2=100+0.176$

$\\ \tt\hookrightarrow v^2=100.176$

$\\ \tt\hookrightarrow v=10.001m/s$

8 0

2 years ago

The period of an oscillating particle is 32 s, and its amplitude is 15 cm. at t = 0, it is at its equilibrium position. find the

morpeh [17]

At t=0, the particle was at its equilibrium position. The time period is 32 seconds, so in 8 seconds, it will reach the extreme location once, and hence, in 8 seconds, it will cover a distance equivalent to its amplitude.

5 0

3 years ago