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3 years ago

Which of the following will not reduce water pollution?

Physics

2 answers:

Rina8888 [55]3 years ago

7 0

Answer:

A. Using artificial fertilizers instead of composting

Explanation:

Composting is actually more better than fertilizers usually, as it takes natural waste and with help of nature, returning the nutrients into soil once again. However, Artificial fertilizers usually uses chemicals to ensure that weed does not grow, and may contain other "unnatural" ingredients to ensure that only the crop grows on the field. Because of the usage of non-natural ingredients, any run-off can prove to be detrimental to any animals that eat any of the run-off infected material, or drink the polluted water.

Scrat [10]3 years ago

7 0

Explanation: Using artificial fertilizers instead of composting is one of the main causes of water pollution. When it rains the artificial fertiliser will run into lakes, ponds, or any other body of water. The water will then be polluted with chemicals. Composting will not do this. There are no chemicals in compost.

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hey my friend rly needed help and i wish i could help do you guys kn what .1895 g of salt per 50 grams of sea water what is its

iragen [17]

Answer:

1 cm3 is = 1 ml. Therefore 1000 g of seawater = 973.71 mL.

Explanation:

Seawater salinity will vary from place to place and with the temperature of the seawater. Of course the composition of dissolved substances in seawater, along with salt that is, will also vary from place to place.

On average, seawater in the world's oceans has a salinity of approximately 3.5%, or 35 parts per thousand. This means that for every 1 litre (1000 mL) of seawater, there are 35 grams of salts (mostly, but not entirely, sodium chloride) dissolved in it.

Seawater has an average density of 1.027 g/cm3, but this varies with temperature and salinity over a range of about 1.020 to 1.029.

5 0

4 years ago

Find the magnitude of the resultant force and the angle it makes with the positive x-axis. (Let |a| = 22 lb and |b| = 16 lb. Rou

SVEN [57.7K]

Incomplete question as the angle between the force is not given I assumed angle of 55°.The complete question is here

Two forces, a vertical force of 22 lb and another of 16 lb, act on the same object. The angle between these forces is 55°. Find the magnitude and direction angle from the positive x-axis of the resultant force that acts on the object. (Round to one decimal places.)

Answer:

Resultant Force=33.8 lb

Angle=67.2°

Explanation:

Given data

Fa=22 lb

Fb=16 lb

Θ=55⁰

To find

(i) Resultant Force F

(ii)Angle α

Solution

First we need to represent the forces in vector form

$\sqrt{x} F_{1}=22j\\ F_{2}=u+v\\F_{2}=16sin(55)i+16cos(55)j\\F_{2}=16(0.82)i+16(0.5735)j\\F_{2}=13.12i+9.176j$

Total Force

$F=F_{1}+F_{2}\\ F_{2}=22j+13.12i+9.176j\\F_{2}=13.12i+31.176j$

The Resultant Force is given as

$|F|=\sqrt{x^{2} +y^{2} }\\|F|=\sqrt{(13.12)^{2} +(31.176)^{2} }\\ |F|=33.8lb$

For(ii) angle

We can find the angle bu using tanα=y/x

$tan\alpha =\frac{31.176}{13.12}\\ \alpha =tan^{-1} (\frac{31.176}{13.12})\\\alpha =67.2^{o}$

7 0

4 years ago

Un bombero alejado d = 31.0 m de un edificio en llamas dirige un chorro de agua desde una manguera contra incendios a nivel del

Mila [183]

Answer:

El chorro golpea el edificio a una altura de 15.943 metros con respecto al suelo.

Explanation:

El chorro de agua exhibe un movimiento parabólico, dado que este tiene una inclinación inicial y la única aceleración es debida a la gravitación terrestre. Las ecuaciones cinemáticas que modelan el fenómeno son:

Distancia horizontal (en metros)

$x = x_{o} + v_{o}\cdot t \cdot \cos \theta$

Donde:

$x_{o}$ - Posición horizontal inicial, medida en metros.

$t$ - Tiempo, medido en segundos.

$v_{o}$ - Velocidad inicial, medida en metros por segundo.

$\theta$ - Angulo de inclinación del chorro de agua, medido en grados sexagesimales.

Distancia vertical (en metros)

$y = y_{o} + v_{o}\cdot t \cdot \sin \theta + \frac{1}{2}\cdot g \cdot t^{2}$

Donde:

$y_{o}$ - Posición vertical inicial, medida en metros.

$g$ - Constante gravitacional, medida en metros por segundo al cuadrado.

Partiendo de la primera ecuación, se despeja el tiempo:

$t = \frac{x - x_{o}}{v_{o}\cdot \cos \theta}$

Si $x = 31\,m$ , $x_{o} = 0\,m$ , $v_{o} = 40\,\frac{m}{s}$ and $\theta = 33^{\circ}$ , entonces:

$t = \frac{31\,m-0\,m}{\left(40\,\frac{m}{s} \right)\cdot \cos 33^{\circ}}$

$t = 0.924\,s$

La altura máxima se calcula por sustitución directa de términos en la segunda ecuación. Si $y_{o} = 0\,m$ , $v_{o} = 40\,\frac{m}{s}$ , $t = 0.924\,s$ y $g = -9.807\,\frac{m}{s^{2}}$ , entonces:

$y = 0\,m + \left(40\,\frac{m}{s} \right)\cdot (0.924\,s)\cdot \sin 33^{\circ} + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (0.924\,s)^{2}$