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kkurt [141]
4 years ago
10

What are the strengths and weaknesses of the four methods of waste management?​

Physics
1 answer:
Nitella [24]4 years ago
4 0

Answer & Explanation: Waste management are all activities and actions required to manage waste from its inception to its final disposal. There are several methods of managing waste with its strengths and weaknesses. The strengths include;

* It creates employment

* It keeps the environment clean

* The practice is highly lucrative

* It saves the earth and conserves energy

The weaknesses of the methods of waste management includes;

* The sites are often dangerous

* The process is mostly

* There is a need for global buy-in

* The resultant product had a short life

You might be interested in
Consider two carts, of masses m and 2m, at rest on an air track. if you first push one cart for 3 s and then the other for the s
Marrrta [24]
F = m₁ a₁ = m₂ a₂

if m₁ = m and m₂ = 2m :

F = ma₁ = 2m  a₂ ⇒ a₁ = 2 a₂
 
since v = at + v₀  with t = 3, v₀ = 0 ⇒ v = 3a:

v₁ = 2 v₂

since p = vm with v₁ = 2v and v₂ = v :

p₁ = v₁m₁ = 2v ⁻ m
p₂ = v₂m₂ = v ⁻ 2m

p₁ = p₂

8 0
3 years ago
One significant result of the Michelson experiment of 1880 was that it:
Alika [10]

Answer:

calculated the speed of light over a short distance

Explanation:

Albert Michelson <u>was an American physicist who conducted an experiment regarding the speed of light in the air 1880s.</u> He believed in the existence of "aether," a field of space that is deemed necessary for transporting electromagnetic forces. In order to do this, he setup a device in order to accurately measure the speed of light in in aether. His device is now known as <u>"Michelson interferometer." </u><em>The result was actually negative. </em>

In order to make his experiment even more accurate, he collaborated with Edward Morley, a famous American scientist. Although the result failed, together, they were able to improve and come with the a standard length of light. This calculated the speed of light over a short distance, which was the significant result of Michelson's experiment.

Thus, this explains the answer.

8 0
3 years ago
Read 2 more answers
Water (rhoH20 = 1000.0 kg/m3 ) flows through a garden hose that goes up a step 20.0 cm high. The cross-sectional area of the hos
Soloha48 [4]

Answer:

 P₂ = 138.88 10³ Pa

Explanation:

This is a problem of fluid mechanics, we must use the continuity and Bernoulli equation

Let's start by looking for the top speed

        Q = A₁ v₁ = A₂ v₂

We will use index 1 for the lower part and index 2 for the upper part, let's look for the speed in the upper part (v2)

         v₂ = A₁ / A₂ v₁

They indicate that A₂ = ½ A₁ and give the speed at the bottom (v₁ = 1.20 m/s)

         v₂ = 2  1.20

         v₂ = 2.40 m / s

Now let's write the Bernoulli equation

        P₁ + ½ ρ v₁² + ρ g y₁ = P2 + ½ ρ v₂² + ρ g y₂

Let's clear the pressure at point 2

       P₂ = P₁ + ½ ρ (v₁² - v₂²) + ρ g (y₁-y₂)

we put our reference system at the lowest point

        y₁ - y₂ = -20 cm

Let's calculate

       P₂ = 143 10³ + ½ 1000 (1.20² - 2.40²) + 1000 9.8 (-0.200)

       P₂ = 143 103 - 2,160 103 - 1,960 103

       P₂ = 138.88 10³ Pa

3 0
4 years ago
A charge partides round a 1 m radius circular particle accelerator at nearly the speed of light. Find : (a) The period (b) The c
Scrat [10]

Explanation:

It is given that,

Radius of circular particle accelerator, r = 1 m

The distance covered by the particle is equal to the circumference of the circular path, d = 2πr

d = 2π × 1 m

(a) The speed of satellite is given by total distance divided by total time taken as :

speed=\dfrac{distance}{time}

Let t is the period of the particle.

t=\dfrac{d}{s}

d = distance covered

s = speed of particle

It is given that the charged particle is moving nearly with the speed of light

t=\dfrac{d}{c}

t=\dfrac{2\pi\times 1\ m}{3\times 10^8\ m/s}

t=2.09\times 10^{-8}\ s

(b) On the circular path, the centripetal acceleration is given by :

a=\dfrac{c^2}{r}

a=\dfrac{(3\times 10^8\ m/s)^2}{1\ m}

a=9\times 10^{16}\ m/s^2

Hence, this is the required solution.

8 0
3 years ago
What is the upper block's acceleration if the coefficient of kinetic friction between the block and the table is 0.13?
IgorLugansk [536]

Let us assume that pulley is mass less.

Let the tension produced at both ends of the pulley is T.

We are asked to calculate the acceleration of the block.

Let the masses of two bodies are denoted as m_{1} \ and\ m_{2}\ respectively

Let\ m_{1} =1 kg\ and\ m_{2} =2 kg

As per this diagram, the body having mass 1 kg is moving downward and the body having mass 2 kg is moving on the surface of the table.

Let the acceleration of each block is a .

For body having mass 1 kg:

The net force acting on 1 kg body will be-

                             m_{1} g-T=m_{1} a        [1]

Here tension in the rope will be vertically upward and weight of the body will be in vertical downward direction.

For body having mass 2 kg:

The coefficient of kinetic friction [\mu]=0.13

Hence\ the\ frictional\ force\ F=\mu N

                                                     F=\mu m_{2} g

Hence the net force acting on the body having mass 2 kg-

                                  T-\mu m_{2} g=m_{2} a  [2]

Here the tension of the rope is towards right i.e along the direction of motion of the 2 kg block and frictional force is towards left.

Combining 1 and 2 we get-

                           m_{1} g-T=m_{1}a             [1]

                           T-\mu m_{2}g= m_{2} a   [2]

                           ---------------------------------------------------

                           [m_{1} -\mu m_{2} ]g=[m_{1} +m_{2} ]a

                           a=\frac{m_{1}-\mu m_{2}} {m_{1}+ m_{2}}*g

                           a=\frac{1-[2*0.13]}{1+2} *9.8\ m/s^2

                           a=\frac{0.74}{3} *9.8\ m/s^2

                           a=2.417 m/s         [ans]

6 0
3 years ago
Read 2 more answers
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