Ask question

Ask question

All categories

photoshop1234 [79]

3 years ago

13

A swimming pool is 25.0 ft. long, 18.5 ft. wide, and 9.0 ft. deep. When filled, the water level is 7.0 inches from the top. Disi

nfectant instructions say to add 18 mg of Cl per liter of water. How many pounds of chlorine are added to the pool?

1 answer:

GarryVolchara [31]3 years ago

8 0

Answer:

1.9841256 kg

Explanation:

Given;

Length of the swimming pool = 25.0 ft = 7.62 m ( 1 ft = 0.3048 m )

Width of the swimming pool = 18.5 ft = 5.64 m

Depth of the pool = 9.0 ft =

Total depth of the water in the pool when filled = 9 ft - 7 inches = 2.56 m

now,

Volume of the water in the pool = Length × Width × Depth

or

Volume of the water in the pool = 7.62 × 5.64 × 2.56 = 110.2292 m³

also,

1 m³ = 1000 L

thus,

110.2292 m³ = 110229.2 L

also it is given that 18 mg of Cl is added to 1 liter of water

therefore,

In 110229.2 L of water Cl added will be = 110229.2 × 18 = 1984125.6 mg

or

= 1.9841256 kg

You might be interested in

A child is holding a wagon from rolling straight back down in a driveway that inclined at 20 degree horizontal. if the wagon wei

Alenkinab [10]

Answer:

F = 51.3°

Explanation:

The component of weight parallel to the inclined plane must be responsible for the rolling back motion of the car. Hence, the force required to be applied by the child must also be equal to that component of weight:

$F = Parallel\ Component\ of\ Weight\ of\ Wagon= WSin\theta\\$

where,

W = Weight of Wagon = 150 N

θ = Angle of Inclinition = 20°

Therefore,

$F = (150\ N)Sin\ 20^o$

<u>F = 51.3°</u>

8 0

2 years ago

Suzy drops a rock from the roof of her house. Mary sees the rock pass her 2.9 m tall window in 0.134 sec. From how high above th

timama [110]

Answer:

Explanation:

Given

length of window $h=2.9\ m$

time Frame for which rock can be seen is $\Delta t=0.134\ s$

Suppose h is height above which rock is dropped

Time taken to cover $h+2.9 is t_1$

so using equation of motion

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

where y=displacement

u=initial velocity

a=acceleration

t=time

time taken to travel h is

$h=0+0.5\times g\times (t_2)^2---2$

Subtract 1 and 2 we get

$2.9=0.5g(t_1^2-t_2^2)$

$5.8=g(t_1+t_2)(t_1-t_2))$

and from equation $t_1-t_2=0.134\ s$

so $t_1+t_2=\frac{5.8}{9.8\times 0.134}$

$t_1+t_2=4.416\ s$

and $t_1=t_2+\Delta t$

so $t_2+\Delta t+t_2=4.416$

$2t_2+0.134=4.416$

$t_2=0.5\times 4.282$

$t_2=2.141\ s$

substitute the value of $t_2$ in equation 2

$h=0.5\times 9.8\times (2.141)^2$

$h=22.46\ m$

8 0

3 years ago

Does the sun always shine directly overhead at the equator at noon

Nikitich [7]

The sun always shines directly overhead at noon. This is because the equator always gets the equivalent amount of sunlight. The area always get 12 hours of sunlight, because it's 0 degrees north and south and it's at the center of the Earth.

4 0

3 years ago

A student is performing experiments on a particular substance. Which

boyakko [2]

Answer: A.

Explanation:

6 0

2 years ago

A 2.2 kgkg block slides along a frictionless surface at 1.2 m/sm/s . A second block, sliding at a faster 4.0 m/sm/s , collides w

aleksklad [387]

Answer:

0.6kg

Explanation:

the unknown here is the mass of the second block

applying the law of the conservation of momentum

m₁v₁ + m₂v₂ = (m₁ + m₂) v₃

where m₁=mass of first block=2.2kg

m₂=mass of colliding block= ?

v₁= velocity of first block=1.2m/s

v₂=velocity of colliding block=4.0m/s

v₃= final velocity of combined block=1.8m/s

applying the formula above

(2.2 × 1.2) + (m₂ × 4) = (2.2 + m₂) × 1.8

2.64 + 4m₂ = 3.96 + 1.8m₂

collecting like terms

4m₂ - 1.8m₂ = 3.96 - 2.64

2.2m₂=1.32

divide both sides by 2.2

m₂= 0.6kg

4 0

3 years ago